Question

A block of mass 1 kg is pushed up a surface inclined to horizontal at an angle of 60° by a force of 10 N parallel to the inclined surface as shown in figure. When the block is pushed up by 10 m along inclined surface, the work done against frictional force is: [g = 10 m/s²]

• A. 10 J
• B. 5×10³ J
• C. 5 J
• D. 5√3 J

Answer 1

Answer: (C)

5 J

Answer 2

The normal force on the block is $N = mg \cos(60^\circ)$. Given $m=1$ kg, $g=10$ m/s², $\theta=60^\circ$. $N = 1 \times 10 \times \frac{1}{2} = 5$ N. The coefficient of kinetic friction is given as $\mu_s = 0.1$, assume $\mu = 0.1$. The frictional force is $F_f = \mu N = 0.1 \times 5 = 0.5$ N. The block is pushed up by $d=10$ m. The work done against frictional force is $W_f = F_f \times d$. $W_f = 0.5 \text{ N} \times 10 \text{ m} = 5$ J.