Question

A 19.7 kg block is dragged over a rough, horizontal surface by a constant force of 188 N acting at an angle of 28.6 ◦ above the horizontal. The block is displaced 48.7 m, and the coefficient of kinetic friction is 0.103.Find the work done by the 188 N force. The acceleration of gravity is 9.8 m/s 2 . Answer in units of J Find the magnitude of the work done by the force of friction. Answer in units of J. a) What is the change in kinetic energy of the crate? Answer in units of J. b) What is the speed of the crate after it is pulled the 9.28 m? Answer in units of m/s.

Answer 1

a) Change in Kinetic energy = Work Done

We require Normal Force N,

N + 188 X sin(28.6 ) - m x g = 0

 so N = m x g - 188 x sin(28.6)  

          = 19.7 Kg x 9.8m/s^2  - 188 x sin(28.6)  

           = 19.7 Kg x 9.8m/s^2  - 188 x 0.478

          = 193.06 Kg.m/s^2 - 89.864 N

          = 103.196 N

Wvertical = F*x = μ X 103.196 x 48.7 = 0.103 X 103.196 x 48.7

= 517.641 N

Whorizontal  = F*x = F x cos(28.6 ) x 48.7  

= 188 x cos(28.6 ) x 48.7

= 188 x 0.877 x 48.7

= 8029.46 J

b)  V^2 = U^2 + 2 x a x s

V =  (2 x a x s)^0.5

= 13.486 m/s

Answer 2

Answer:

Part a)

$W = 8038.5 J$

Part b)

$W_f = -516.6 J$

Part c)

$\Delta K = 7521.9 J$

Part d)

$v_f = 27.6 m/s$

Explanation:

As per force equation on the block we have

$mg = F_n + F sin\theta$

$19.7 \times 9.8 = F_n + 188 sin28.6$

$F_n = 103 N$

Now we have

$F_x - F_f = F_{net}$

$188 cos28.6 - \mu F_n = ma$

$165 - (0.103)(103) = 19.7 a$

$a = 7.84 m/s^2$

Part a)

Now we have to find work done by applied force

$W = F.d$

$W = F d cos\theta$

$W = 188 \times 48.7 cos28.6$

$W = 8038.5 J$

Part b)

Now work done by friction force is given as

$W_f = -\mu F_n .d$

$W_f = -(0.103)(103)(48.7)$

$W_f = -516.6 J$

Part c)

By work energy theorem we have

Total work done = change in kinetic energy

$8038.5 - 516.6 = W_{net}$

$W_{net} = 7521.9 J$

$\Delta K = 7521.9 J$

Part d)

Final speed of the block is given as

$\frac{1}{2}mv_f^2 = 7521.9$

$\frac{1}{2}(19.7)v_f^2 = 7521.9$

$v_f = 27.6 m/s$