Question

An 8.5-kg block is pushed along a horizontal rough surface by a 40-N force inclined at
20° with the horizontal. The coefficient of friction between the surface and block is 0.35.
If the block has an initial velocity of 3.6 m/s and the force does 200 J of work on the
block, find:
(a) The total distance moved by the block.
(b) The final velocity of the block.

Answer 1

Given:-

Mass of the block m = 8.5kg

Force F = 40N.

θ = 20°

Coefficient of friction μ =0.35

Initial Velocity of block v = 3.6m/s.

Work done by force F, W = 200J.

To Find:-

a) The total distance moved by the block

b) The final velocity of the block

Solution:-

a) Work Done by the force

W = Fdcosθ

200 = 40xdcos20°

By solve we get d = 5.32m

b)  From the free body diagram ( Find attached figure)

$F_{net} = Fcos\theta - F_{f}$

$F_{f}$ = Friction Force

$F_{f}$ = μN

then , N = mg + Fsinθ

N = 8.5x9.81 + θ0sin20°

N = 97.06N

$F_{f}$ = 0.35x97.06N

$F_{f}$ = 33.973N

$F_{net}$ = 40cos20° - 33.973

$F_{net}$ = 3.6146

As per Newton's law

$F_{net}$ = m x  acceleration(a)

a = $\frac{3.646}{8.5}$

a = 0.4252m/s²

We have

v² = u² + 2as

v² = (3.6)² + 2(0.4252)(5.32)        {s= d = 5.32}

v² = 12.96 + 4.524128

v² = 17.4841

v = 4.181m/sec