Can someone solve 64 A block of mass 500 g is attached to a spring of spring con
ID: 3278284 • Letter: C
Question
Can someone solve 64 A block of mass 500 g is attached to a spring of spring constant 80 N/m (see the following figure). The other end of the spring is attached to a support while the mass rests on a rough surface with a coefficient of friction of 0.20 that is inclined at angle of 30 degree The block is pushed along the surface till the spring compresses by 10 cm and is then released from rest. (a) How much potential energy was stored in the block-spring-support system when the block was just released? (b) Determine the speed of the block when it crosses the point when the spring is neither compressed nor stretched. (c) Determine the position of the block where it just comes to rest on its way up the incline. A block of mass 200 g is attached at the end of massless spring of spring constant 100 N/cm. The other end of the spring is attached to the ceiling and the mass is brought to rest. Let us mark this point as O. Suppose, this point is taken to be the zero of the potential energy of the block, both from the weight and the spring force. The mass hangs freely and the spring is in a stretched state. The block is then pulled downward by another 5.00 cm and released from rest. (a) What is the net potential energy of the block at the instant the block is at lowest point?Explanation / Answer
64. (a) PE stored = k d^2 / 2
= (80)(0.10^2) / 2
= 0.4 J
(b) Applying energy conservation,
Work done by gravity + work done by friction + work done by spring = change in KE
{ f = uk m g cos30 = 0.20 x 0.500 x 9.81 x cos30 = 0.85 }
- m g d sin30 - f d + 0.4 = 0.500 (v^2 - 0) / 2
- (0.500 x 9.81 x 0.10 sin30) - (0.85 x 0.10) + 0.4 =0.250 v^2
0.250 v^2 = 0.07
v = 0.53 m/s ...........Ans
(c) after that ,
- m g d' sin30 - f d' = 0.500 ( 0 - 0.53^2) / 2
- (0.500 x 9.81 x d' sin30) - (0.85 d') = - 0.0698
- 2.45 d' - 0.85 d' = - 0.0698
d' = 0.021 m Or 2.11 cm
from initial position = 10 + 2.11 = 12.11 cm