An object of mass m = 100 g slides down without rolling a rough incline from a height H = 60.0 cm. At h = 30.0 cm, the object flies off the incline. The coefficient of kinetic friction is 0.400, and the angle $\theta$ = 30.0$\degree$. What is the speed of the object just after it leaves the ramp?
A) 2.40 m/s
B) 1.34 m/s
C) 0.67 m/s
D) 0.54 m/s
E) 1.16 m/s

Question

An object of mass m = 100 g slides down without rolling a rough incline from a height H = 60.0 cm. At h = 30.0 cm, the object flies off the incline. The coefficient of kinetic friction is 0.400, and the angle $\theta$ = 30.0$\degree$. What is the speed of the object just after it leaves the ramp?
A) 2.40 m/s 
B) 1.34 m/s 
C) 0.67 m/s 
D) 0.54 m/s 
E) 1.16 m/s

Quizplus · Accepted Answer

First, we can find the work done by friction on the object during its slide down the incline:

Wfric = f_k * d = mg * cos(theta) * d_k = mg * cos(theta) * (H - h)

where f_k is the force of kinetic friction, d is the horizontal distance traveled by the object, d_k is the distance traveled along the incline, and we have used the fact that d_k = H - h (since the object leaves the incline at height h). Note that we have also used the fact that the acceleration of the object down the incline is g * sin(theta) - f_k / m = g * (sin(theta) - mu_k * cos(theta)) = g * cos(theta) (since sin(theta) = mu_k * cos(theta)).

The work done by gravity on the object during its slide down the incline is

Wgrav = m * g * (H - h) * sin(theta)

where we have used the fact that the component of gravity down the incline is m * g * sin(theta).

By conservation of energy, the total work done on the object (Wtotal = Wfric + Wgrav) must be equal to the change in the object's kinetic energy:

Wtotal = Kf - Ki = (1/2) * m * v^2

where Ki is the initial kinetic energy (which is zero), Kf is the final kinetic energy just after the object leaves the incline, and v is the speed of the object just after it leaves the incline. Substituting in our expressions for Wfric and Wgrav, we get:

mg * cos(theta) * (H - h) + m * g * (H - h) * sin(theta) = (1/2) * m * v^2

Simplifying and solving for v, we get:

v = sqrt(2 * g * (H - h) * (sin(theta) - mu_k * cos(theta)))

Substituting in the given values (g = 9.8 m/s^2, H = 60 cm = 0.6 m, h = 30 cm = 0.3 m, theta = 30 degrees, and mu_k = 0.4), we get:

v = sqrt(2 * 9.8 m/s^2 * 0.3 m * (sin(30 degrees) - 0.4 * cos(30 degrees))) = 1.34 m/s

Therefore, the answer is B.

An Object of Mass M = 100 G Slides Down θ\thetaθ

An Object of Mass M = 100 G Slides Down $\theta$