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
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
$(\mathrm{W})=\Delta \mathrm{K}$
$\Delta \mathrm{K} =\mathrm{W}-\mathrm{W}_{2}-\Delta \mathrm{U}$
$=100-8.66-50$
$=41.34 \mathrm{~J}$
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