Question 5

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 s$$^{-2}$$

Frictional force acting on an inclined plane: $$f_k = \mu_k N = \mu_k mg \cos\theta$$

$$f_k = 0.1 \times 1 \times 10 \times \cos(60^\circ) = 1 \times \frac{1}{2} = 0.5\text{ N}$$

$$W_{\text{against}} = f_k \cdot d$$ $$\implies W_{\text{against}} = 0.5 \times 10 = 5\text{ J}$$

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