The question is not complete and the complete question is;
A moving 1.60 kg block collides with a horizontal spring whose spring constant is 262 N/m.
The block compresses the spring a maximum distance of 13.00 cm from its rest position. The coefficient of kinetic friction between the block and the horizontal surface is 0.260. A)What is the work done by the spring in bringing the block to rest?
B)How much mechanical energy is being dissipated by the force of friction while the block is being brought to rest by the spring?
C)What is the speed of the block when it hits the spring?
Answer:
A) Work done by spring = 1.31 J
Step-by-step explanation:
We are given that;
Mass(m) = 1.6 kg
Spring Constant (k) = 262 N/m
Maximum distance of compression (x) = 13 cm = 0.13m
Coefficient of kinetic friction = 0.26
Now, from Hooke's law, the work done by the spring is given by;
W = (1/2)k•x²
Plugging in the relevant values, we obtain ;
W = (1/2) x 262 x 0.13² = 1.31 J
B) The energy dissipated by the force of friction which is the work done by the friction is given by the product of the friction force and the displacement of the block.
Thus, W_f = F_f•x
Where, F_f is frictional force which is given by, F_f = μmg
μ = coefficient of kinetic friction = 0.26
F_f = 0.26 x 1.6 x 9.81 = 4.08 N
Thus,
W_f = 4.08 x 0.13 = 0.53 J
C) Now, we can find the velocity by applying conservation of energy;
Thus,
(1/2)mv² = W + W_f
(1/2)mv² = 1.31 + 0.53
(1/2)mv² = 1.84
mv² = 1.84 x 2
v² = (1.84 x 2)/1.6 = 2.3
Thus,v = √2.3 = 1.52 m/s