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  • A block of mass 1 kg is pushed up a surface inclined to horizontal at an angle of 30^o by a force of 10 N parallel to the inclined surface.

A block of mass 1 kg is pushed up a surface inclined horizontally at an angle of30^{\circ} a force of 10 N parallel to the inclined surface. The coefficient of friction between block and the incline is 0.1. If the block is pushed up by 10 m along the incline, calculate

a) work done against gravity

b) work done against the force of friction

c) increase in potential energy

d) increase in kinetic energy

e) work done by an applied force

Answers (1)

Mass = 1 kg, \theta = 30, F = 10 N, d = 10m

(a) work done against gravity = mgh

\sin 30 = \frac{h}{10}

h= \frac{10}{2}=5m

Work done = 1 \times 10 \times 5 = 50 J

(b) Work done against the force of friction = \mu f.s = \mu mg \cos \theta \times s

= 0.1 \times 1 \times 10 \cos 30 \times 10 = 10 \times \frac{\sqrt{3}}{2} = 5 \sqrt{3} Joules

(c) increase in potential energy

This value is equal to the work done against gravity, which we have calculated above in part a. hence, the increase in potential energy = 50J

(d) increase in kinetic energy

\Delta KE= work done = -mgh + F_s - f_s

= - 50 - 5 \sqrt{3 }+ 10 \times 10 = 50 - 5\sqrt{ 3}

= 41.340 J

(e) work done by applied force.

= F.S = 10 \times 10 = 100 J

Posted by

infoexpert24

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