A solid block of mass 25.0 kg starts from rest and slides down a rough surface w

Question

A solid block of mass 25.0 kg starts from rest and slides down a rough surface with a length of 10.0 m where coefficient of friction is mu_k = 0.400. The surface is inclined at 50.0 degree. After sliding 10.0 meters down the incl. it slides onto a flat surface with kinetic coefficient of friction of 0200 moves 5.00 m. It then hits a spring wit spring constant of k_s = 300.0 N/m Draw a freebody Diagram indicating all forces. Write out and solve the force and energy equations. How high is the block the flat surface at the star? Solve for the normal force Solve for the kinetic friction on the block Solve for F_x = mg sin theta - f_k Solve for the acceleration down the incline of the block What is the change in the internal energy of the system due to friction after block has moved 5.00 m? What is the kinetic energy of the block after block has moved 5.00 m?? How much is the spring compressed by the block? How much spring potential energy does the spring have at this p

Explanation / Answer

1. on flat surface:

N = m g and f = uk m g

change in internal energy = work done by friction

= f.d = f d cos180

= - uk m g d

= - 0.2 x 25 x 9.8 x 5

= - 245 J ........Ans

2. for KE at bottom of incline,

v^2 - u^2 = 2 a d

v^2 - 0 = 2 x 4.99 x 10 = 99.8

KEi = m v^2 /2 = 1247.5 J

after block has moved 5 m,

KEf = 1247.5 + ( -245) = 1002.5 J ........Ans

3. Applying work energy theorem,

work done by spring + work done by friction = change in KE

- k x^2 /2 - (0.2 * 25 * 9.8 * x) = 0 - 1002.5

150x^2 + 49x - 1002.5 = 0

x = 2.43 m

4. Spring PE = k x^2 /2

= 300 x (2.43^2) / 2

= 885.7 J

5. after compression,

KE of block at bottom of incline = 885.7 - (0.2 x 25 x 9.8 x (5 + 2.43))

= 521.63 J

now force along incline = ( (0.4 x 25 x 9.8 x cos50) + (25 x 9.8 x sin50))

= 250.67

Applying work - energy theorem,

- 250.67 L = 0 - 521.63

L = 2.08

block moves up the incline 2.08 m along the incline.

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