Question

A block of mass 5.6 kg is attached to a horizontal spring on a rough floor. Initially the spring is compressed 3.5 cm. The spring has a force constant of 1040 N/m. The coefficient of kinetic friction between the block and the floor is 0.26. The block is released from rest. The speed of the block after it has traveled 0.020 m is

Answer 1

Answer:

The velocity of block = 0.188 $\frac{m}{s}$

Explanation:

Mass m = 5.6 kg

k = 1040 $\frac{N}{m}$

$\mu$ = 0.26

$x_{1} =$ 0.035 m  , $v_{1}$ = 0

$x_{2} =$ 0.02 m

From work energy theorem

$K_{1} + U_{1} + W_{other} = K_{2} + U_{2}$  --------- (1)

Kinetic energy

$K = \frac{1}{2} k x^{2}$  ------- (1)

Potential energy

$U = \frac{1}{2} k x^{2}$ ------- (2)

Work done

W = F.s ------ (3)

From Newton's second law

$R_{N}$ = mg

$R_{N}$  = 5.6 × 9.81 = 54.9 N

Friction  force = 0.4 × 54.9 = 21.9 N

Now the work done by the friction

$W_{f}$ = - 21.9 × 0.015

$W_{f}$ = - 0.329 J

Now kinetic energy

At point 1

$K_{1} = \frac{1}{2} m v^{2} _{1}$

$K_{1} = 0$

$U_{1} = \frac{1}{2} k x^{2}$

$U_{1} = \frac{1}{2} (1040) 0.035^{2}$

$U_{1} =$ 0.637 J

At point 2

$K_{2} = \frac{1}{2} (5.6) v^{2} _{2}$

$K_{2} = 2.8 v_{2} ^{2}$

Potential energy

$U_{2} = \frac{1}{2} k x_2^{2}$

$U_{2} = \frac{1}{2} (1040) 0.02^{2}$

$U_{2} = 0.208$ J

From equation (1) we get

0 + 0.637 - 0.329 = 2.8 $v_{2} ^{2}$ + 0.208

2.8 $v_{2} ^{2}$ = 0.1

$v_{2} =$ 0.188 $\frac{m}{s}$

This is the velocity of block.