An object with a mass of 5.5 kg is allowed to slide from rest down an inclined plane. The plane makes an angle of 30^o with the horizontal and is 72 m long. The coefficient of friction between the plane and the object is 0.35. The speed of the object at the bottom of the plane is A) 5.3 m/s. B) 15 m/s. C) 24 m/s. D) 17 m/s. E) 11 m/s.

Question

Quizplus · Accepted Answer

First, we need to find the acceleration of the object down the plane using the equation:

a = gsinθ - μgcosθ

where g is the acceleration due to gravity (9.8 m/s2), θ is the angle of the plane (30°), and μ is the coefficient of friction (0.35).

a = (9.8 m/s2)sin(30°) - (0.35)(9.8 m/s2)cos(30°)
a = 4.53 m/s2

Next, we can use the equation for acceleration to find the final velocity of the object at the bottom of the plane:

v^2 = u^2 + 2as

where u is the initial velocity (0 m/s), s is the distance down the plane (72 m), and a is the acceleration we just calculated.

v^2 = 2(4.53 m/s2)(72 m)
v = 16.9 m/s

However, this is just the speed without considering the force of friction. We need to adjust for the work done against friction:

Wfriction = μmgd

where m is the mass of the object (5.5 kg) and d is the distance down the plane (72 m).

Wfriction = (0.35)(5.5 kg)(9.8 m/s2)(72 m) = 135.51 J

The work-energy principle states that the work done on an object is equal to its change in kinetic energy. In this case, the work done on the object is:

W = Fd = μmgd

where F is the force of friction. So we can write:

Fd = 1/2mvf^2 - 1/2mu^2

Solving for vf, we get:

vf = sqrt[2Fd/m + u^2]

Substituting in the values we know:

vf = sqrt[2(135.51 J)/(5.5 kg) + 0^2] = 17 m/s

Therefore, the answer is D) 17 m/s.

An Object with a Mass of 5