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 (figure). The coefficient of friction between block and the incline is $0.1$. If the block is pushed up by $10\,m$  along the inclined calculate 

$(a)$ work done against gravity

$(b)$ work done against force of friction

$(c)$ increases in potential energy

$(d)$ increases in kinetic energy

$(e)$ work done by applied force

$m=1 \mathrm{~kg}, \theta=30^{\circ}, \mathrm{F}=10 \mathrm{~N}, \mu=0.1, d=10 \mathrm{~m}$

$(a)$Work done against gravity,

$\mathrm{W}_{1} =m g \times \text { Vertical distance travelled }$

$=m g \times d(\sin \theta)=(m g d) \sin \theta$

$=1 \times 10 \times 10 \sin 30^{\circ}=50 \mathrm{~J} \quad\left(\because g \leq 10 \mathrm{~m} / \mathrm{s}^{2}\right)$

$(b)$ Work done against friction,

$\mathrm{W}_{2} =f \times d=\mu \mathrm{N} \times \mathrm{s}=\mu \mathrm{mg} \cos \theta \times d$

$=0.1 \times 1 \times 10 \times \cos 30^{\circ} \times 10$

$=10 \times 0.866=8.66 \mathrm{~J}$

$\text { (c) Increase in } \mathrm{PE}=m g h=m g(d \sin \theta)$

$\Delta \mathrm{U} =1 \times 10 \times 10 \times \sin 30^{\circ}$

$=100 \times \frac{1}{2}=50 \mathrm{~J}$

$(e)$ Work done by applied force,

$\mathrm{W} =\mathrm{F} d$

$=(10)(10)=100 \mathrm{~J}$

$(d)$ By work-energy theorem, we know that work done by all the forces = change in KE

$(\mathrm{W})=\Delta \mathrm{K}$

$\Delta \mathrm{K} =\mathrm{W}-\mathrm{W}_{2}-\Delta \mathrm{U}$

$=100-8.66-50$

$=41.34 \mathrm{~J}$