Question 1

A block is pushed across a rough horizontal surface from point A to point B by a force (magnitude P = 5.4 N) as shown in the figure. The magnitude of the force of friction acting on the block between A and B is 1.2 N and points A and B are 0.5 m apart. If the kinetic energies of the block at A and B are 4.0 J and 5.6 J, respectively, how much work is done on the block by the force P between A and B?

Accepted Answer

The work done on the block by the force P between A and B is given by:
Work = Force x Displacement x cos(theta)
where theta is the angle between the force and the displacement. In this case, the force and the displacement are in the same direction, so theta = 0. Therefore:
Work = Force x Displacement = 5.4 N x 0.5 m = 2.7 J
However, the work done by the force P is partially converted to kinetic energy of the block, and partially dissipated as heat due to friction. The net work done on the block is equal to the change in kinetic energy:
Net work = Kinetic energy at B - Kinetic energy at A = 5.6 J - 4.0 J = 1.6 J
Therefore, the work done by the force P that is converted to kinetic energy is:
Work converted = Net work - Work lost to friction = 1.6 J - 1.2 J = 0.4 J
Since Work = Force x Displacement, we can calculate the magnitude of the displacement from A to B as:
Displacement = Work / Force = 0.4 J / 5.4 N = 0.074 m
Therefore, the average force exerted by the block during the displacement is:
Average force = Work / Displacement = 0.4 J / 0.074 m = 5.4 N (approximately)
This is the same as the magnitude of the force P, which confirms that our calculations are correct.

Question 2

A 2.5-kg object falls vertically downward in a viscous medium at a constant speed of 2.5 m/s. How much work is done by the force the viscous medium exerts on the object as it falls 80 cm?

Accepted Answer

D

Question 3

The horizontal surface on which the block slides is frictionless. The speed of the block before it touches the spring is 6.0 m/s. How fast is the block moving at the instant the spring has been compressed 15 cm? k = 2.0 kN/m

Accepted Answer

A

Question 4

A force acting on an object moving along the x axis is given by Fx = (14x - 3.0x2) N where x is in m. How much work is done by this force as the object moves from x = -1 m to x = +2 m?

Accepted Answer

The answer of A force acting on an object moving...

Question 5

A 1.5-kg object moving along the x axis has a velocity of +4.0 m/s at x = 0. If the only force acting on this object is shown in the figure, what is the kinetic energy of the object at x = +3.0 m?

Accepted Answer

The answer of A 1.5-kg object moving along the x...

Question 6

A 4.0-kg block is lowered down a 37$\degree$ incline a distance of 5.0 m from point A to point B. A horizontal force (F = 10 N) is applied to the block between A and B as shown in the figure. The kinetic energy of the block at A is 10 J and at B it is 20 J. How much work is done on the block by the force of friction between A and B?

Accepted Answer

The answer of A 4.0-kg block is lowered down a...

Question 7

The only force acting on a 2.0-kg body moving along the x axis is given by Fx = (2.0x) N, where x is in m. If the velocity of the object at x = 0 is +3.0 m/s, how fast is it moving at x = 2.0 m?

Accepted Answer

The answer of The only force acting on a 2.0-kg...

Question 8

How much work is done by a person lifting a 2.0-kg object from the bottom of a well at a constant speed of 2.0 m/s for 5.0 s?

Accepted Answer

The work done is calculated using the formula Work = Force x Distance. Since the object is lifted at a constant speed, the force exerted equals the weight of the object (mass x gravity). Here, gravity (g) is 9.8 m/s^2. So, Force = 2.0 kg x 9.8 m/s^2 = 19.6 N. The distance (d) the object is lifted can be found by multiplying the speed (2.0 m/s) by the time (5.0 s), giving d = 10.0 m. Therefore, Work = 19.6 N x 10.0 m = 196 Joules or 0.196 kJ, which rounds to 0.20 kJ.

Question 9

A block slides on a rough horizontal surface from point A to point B. A force (magnitude P = 2.0 N) acts on the block between A and B, as shown. Points A and B are 1.5 m apart. If the kinetic energies of the block at A and B are 5.0 J and 4.0 J, respectively, how much work is done on the block by the force of friction as the block moves from A to B?

Accepted Answer

The answer of A block slides on a rough horizontal...

Question 10

An object moving along the x axis is acted upon by a force Fx that varies with position as shown. How much work is done by this force as the object moves from x = 2 m to x = 8 m?

Accepted Answer

The answer of An object moving along the x axis...

Question 11

A body moving along the x axis is acted upon by a force Fx that varies with x as shown. How much work is done by this force as the object moves from x = 1 m to x = 8 m?

Accepted Answer

The answer of A body moving along the x axis...

Question 12

A 2.0-kg projectile moves from its initial position to a point that is displaced 20 m horizontally and 15 m above its initial position. How much work is done by the gravitational force on the projectile?

Accepted Answer

The answer of A 2.0-kg projectile moves from its initial...

Question 13

A 2.0-kg body moving along the x axis has a velocity vx = 5.0 m/s at x = 0. The only force acting on the object is given by Fx = (-4.0x) N, where x is in m. For what value of x will this object first come (momentarily) to rest?

Accepted Answer

The work done by the force Fx = (-4.0x) N will equal the initial kinetic energy of the object. Solving for x when the kinetic energy is zero gives x = 3.5 m.

Question 14

A 2.0-kg block slides down a frictionless incline from point A to point B. A force (magnitude P = 3.0 N) acts on the block between A and B, as shown. Points A and B are 2.0 m apart. If the kinetic energy of the block at A is 10 J, what is the kinetic energy of the block at B?

Accepted Answer

Since the incline is frictionless, the work done by the force P is equal to the change in kinetic energy of the block. Using the work-energy theorem:

W = ΔK

P × d = Kf - Ki

(3.0 N)(2.0 m) = Kf - 10 J

Kf = 6.0 J + 10 J

Kf = 16 J

Therefore, the kinetic energy of the block at B is 16 J.

Question 15

If the resultant force acting on a 2.0-kg object is equal to $(3 \tilde{\mathbf{i}}+4 \tilde{\mathbf{j}})$ N, what is the change in kinetic energy as the object moves from $(7 \tilde{\mathbf{i}}-8 \tilde{\mathbf{j}})$ m to $(11 \tilde{\mathbf{i}}-5 \tilde{\mathbf{j}})$ m?

Accepted Answer

The answer of If the resultant force acting on a...

Question 16

A 3.0-kg block is dragged over a rough horizontal surface by a constant force of 16 N acting at an angle of 37$\degree$ above the horizontal as shown. The speed of the block increases from 4.0 m/s to 6.0 m/s in a displacement of 5.0 m. What work was done by the friction force during this displacement?

Accepted Answer

The answer of A 3.0-kg block is dragged over a...

Question 17

A 2.0-kg block sliding on a frictionless horizontal surface is attached to one end of a horizontal spring (k = 600 N/m) which has its other end fixed. The speed of the block when the spring is extended 20 cm is equal to 3.0 m/s. What is the maximum speed of this block as it oscillates?

Accepted Answer

The maximum speed of the block occurs when the spring is at its equilibrium position, where all the energy is kinetic. Using the conservation of energy principle, the total mechanical energy at any point during the oscillation is constant. When the spring is extended by 20 cm (0.2 m) and the block's speed is 3.0 m/s, the mechanical energy is a combination of kinetic energy (due to the block's speed) and potential energy (due to the spring's extension). The kinetic energy (KE) is given by $KE = \frac{1}{2}mv^2$ and the potential energy (PE) stored in the spring is given by $PE = \frac{1}{2}kx^2$, where $m$ is the mass of the block, $v$ is the velocity of the block, $k$ is the spring constant, and $x$ is the extension of the spring from its equilibrium position.Given:- $m = 2.0 \, \text{kg}$- $v = 3.0 \, \text{m/s}$ (when $x = 0.2 \, \text{m}$)- $k = 600 \, \text{N/m}$- $x = 0.2 \, \text{m}$The total mechanical energy when the spring is extended by 20 cm is:$$E = \frac{1}{2}mv^2 + \frac{1}{2}kx^2 = \frac{1}{2}(2.0)(3.0)^2 + \frac{1}{2}(600)(0.2)^2 = 9 + 12 = 21 \, \text{J}$$At the equilibrium position, all this energy will be kinetic, so:$$E = \frac{1}{2}mv_{\text{max}}^2$$Solving for $v_{\text{max}}$:$$21 = \frac{1}{2}(2.0)v_{\text{max}}^2$$$$v_{\text{max}}^2 = \frac{21}{1}$$$$v_{\text{max}} = \sqrt{21} \approx 4.58 \, \text{m/s}$$The closest answer to $4.58 \, \text{m/s}$ is 4.6 m/s, which is option A.

Question 18

The only force acting on a 1.8-kg body as it moves along the x axis is given by Fx = -(3.0x) N, where x is in m. If the velocity of the body at x = 0 is vx = +8.0 m/s, at what value of x will the body have a velocity of +4.0 m/s?

Accepted Answer

The answer of The only force acting on a 1.8-kg...

Question 19

A 2.0-kg block situated on a frictionless incline is connected to a light spring (k = 100 N/m), as shown. The block is released from rest when the spring is unstretched. The pulley is frictionless and has negligible mass. What is the speed of the block when it has moved 0.20 m down the plane?

Accepted Answer

The answer of A 2.0-kg block situated on a frictionless...

Question 20

A 10-kg block on a horizontal frictionless surface is attached to a light spring (force constant = 0.80 kN/m). The block is initially at rest at its equilibrium position when a force (magnitude P = 80 N) acting parallel to the surface is applied to the block, as shown. What is the speed of the block when it is 13 cm from its equilibrium position?

Accepted Answer

The answer of A 10-kg block on a horizontal frictionless...

Quiz 6: Work, Force and Energy