Question

A worker pushed a 25.0 kg block 6.50 m along a level floor at constant speed with a force directed 30.0° below the horizontal. If the coefficient of kinetic friction between block and floor was 0.410, what were (a) the work done by the worker's force and (b) the increase in thermal energy of the block-floor system?

Answer 1

Answer:

a) 855.41J

b) 855.41J

Explanation:

The force equation is

N - mg - F sinθ = 0

N = mg + F sinθ

The frictonal  force is =

$F_x = F cos \theta$

$U_xN = Fcos\theta$

substitute the value of the normal force in the above equation

$U_x (mg + Fsin\theta ) = F cos\theta$

$F = \frac{U_x mg}{cos\theta - U_x sin\theta } \\\\= \frac{0.41 \times 25 \times 9.8 }{cos 30^\circ- 0.41 \times sin30 ^\circ}$

= 151.96N

a) work done = F × d

= (151.96) × (6.5) cos30°

= 855.41 J

b) The increase in thermal energy of the block floor system will be equal to the work done by the worker

work done = F × d

= (151.96) × (6.5) cos30°

= 855.41 J