$A$ block of mass $1\,kg$ is pushed up a surface inclined to the horizontal at an angle of $30^{\circ}$ by a force of $10\,N$ parallel to the inclined surface (see figure). The coefficient of friction between the 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 the applied force

(N/A) Given: $m = 1\,kg$,$\theta = 30^{\circ}$,$F = 10\,N$,$\mu = 0.1$,$d = 10\,m$,$g = 10\,m/s^2$.
$(a)$ Work done against gravity $(W_g)$:
$W_g = mgh = mg(d \sin \theta) = 1 \times 10 \times 10 \times \sin 30^{\circ} = 100 \times 0.5 = 50\,J$.
$(b)$ Work done against friction $(W_f)$:
The normal force $N = mg \cos \theta$. The frictional force $f = \mu N = \mu mg \cos \theta$.
$W_f = f \times d = \mu mg \cos \theta \times d = 0.1 \times 1 \times 10 \times \cos 30^{\circ} \times 10 = 10 \times \frac{\sqrt{3}}{2} \approx 8.66\,J$.
$(c)$ Increase in potential energy $(\Delta U)$:
$\Delta U = mgh = mg(d \sin \theta) = 1 \times 10 \times 10 \times 0.5 = 50\,J$.
$(d)$ Increase in kinetic energy $(\Delta K)$:
By the work-energy theorem,$W_{net} = \Delta K$.
$W_{net} = W_{applied} - W_g - W_f = 100 - 50 - 8.66 = 41.34\,J$.
$(e)$ Work done by applied force $(W_{applied})$:
$W_{applied} = F \times d = 10 \times 10 = 100\,J$.