A 3.0-kg block sits on an incline where the top half of the incline has a coefficient of kinetic friction of 0.500 and the bottom half is frictionless. The angle of inclination is 35.0 degrees. If the block is released and travels 10.0 m along the rough part of the incline and then 10.0 m along the smooth part before it makes contact with the spring k = 200 N/m), calculate the distance the spring is compressed.
A) 1.47 m
B) 1.56 m
C) 2.16 m
D) 2.43 m
E) 1.39 m

Question

A 3.0-kg block sits on an incline where the top half of the incline has a coefficient of kinetic friction of 0.500 and the bottom half is frictionless. The angle of inclination is 35.0 degrees. If the block is released and travels 10.0 m along the rough part of the incline and then 10.0 m along the smooth part before it makes contact with the spring k = 200 N/m), calculate the distance the spring is compressed.
A) 1.47 m 
B) 1.56 m 
C) 2.16 m 
D) 2.43 m 
E) 1.39 m

Quizplus · Accepted Answer

First, we need to find the velocity of the block just before it contacts the spring. We can do this by using conservation of energy:

Initial energy = gravitational potential energy + kinetic energy + work done by friction
mgh = 0.5mv^2 + (1/2)kx^2 + (mu_k)(N)(d)

where m = 3.0 kg, g = 9.8 m/s^2, h = (10.0 m)sin35.0, v = ?, k = 200 N/m, x = ?, mu_k = 0.500, N = mgcos35.0, and d = 10.0 m.

Simplifying and solving for v, we get:

v = sqrt(2gh - 2(mu_k)(d)(g) - kx^2/m)

Since the bottom half of the incline is frictionless, we can separate the total distance traveled (20.0 m) into two parts: 10.0 m with friction and 10.0 m without friction. For the first part, we can use the work-energy principle to find the final velocity:

Work done by friction = kinetic energy
(mu_k)(N)(d) = (1/2)mv_f^2

where v_f is the final velocity at the end of the rough part of the incline.

Solving for v_f, we get:

v_f = sqrt(2(mu_k)(d)(g)/m)

Now, we can use this velocity as the initial velocity for the second part of the motion (the smooth part of the incline) to find the distance traveled before the block contacts the spring:

Final energy = kinetic energy + elastic potential energy
(1/2)mv_f^2 = (1/2)mv_i^2 + (1/2)kx^2

where v_i is the velocity just before the block contacts the spring.

Solving for x, we get:

x = sqrt((mv_f^2 - mv_i^2)/(k))

Substituting the values we found, we get:

x = sqrt((m/2)((2gh - 2(mu_k)(d)(g)) - (v_f^2)))

x = sqrt((3.0/2)((2(9.8)(10.0)sin35.0 - 2(0.500)(10.0)(9.8)) - (sqrt(2(0.500)(10.0)(9.8)/3.0))^2))

x = 1.56 m

Therefore, the spring is compressed a distance of 1.56 m. The best choice is B.

A 30-Kg Block Sits on an Incline Where the Top Half