A solid block of mass 70kg starts from rest and slides down a rough surface with

Question

A solid block of mass 70kg starts from rest and slides down a rough surface with a length of 10m where the coefficient of friction is 0.350. The surface is inclined at 50 degrees. After sliding 10 meters down the incline it slides onto a flat surface with a kinetic coefficient of friction of 0.200 moves 5.00m. It then hits a spring with a spring constant of 360 N/m.

1) After the block has slide 10.0m down the incline
a)What is the change in potential energy due to gravity?
b) What is the change in the internal energy of the system due to friction?
c) What is the kinetic energy of the block?
d) What is the speed of the block?
e)How long does it take to slide down?
f) What is the net power of the block?

2) The block slides onto the rough flat surface and hits the spring.
a) What is the change in the internal energy of the system due to friction?
b)What is the kinetic energy of the block after block has moved 5.00m?
c)How much is the spring compressed by the block?
d)How much spring potential energy does the spring have at this point?
e) How far will the block move from the compression point?

Explanation / Answer

1 . a) P.E. due gravity = mglsin = 70 x 9.81 x10 x sin50 =5260.43 J

b) due to friction = mgcos x l = .350 x 70 x 9.81 x10 x cos50 = 1544.91 J

c) K.E. = P.E. due gravity - due to friction = 5260.43 - 1544.91 = 3715.52 J

d) 70 v2 /2 = 3715.52

v = 10.30 m/s

e) 70a = F =mgsin - mgcos = 265.39

a = 3.79 m/s2 downwards

v = u +a t

10.30 = 0+ 3.79t

t = 2.72 sec

f) net power = net work done / time taken = 3715.52 / 2.72 = 1367.65 W

2.a) due to friction = mgl = 0.20 x 70 x 9.81 x 5 = 686.7 J

b) K.E. = 3715.52 - 686.7 = 3028.82 J

c) K.E. = kx2 /2

3028.82 = 360x2 /2

x = 4.10 m

d P.E. = kx2 /2 = 3028.82 J

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