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
$(\mathrm{W})=\Delta \mathrm{K}$
$\Delta \mathrm{K} =\mathrm{W}-\mathrm{W}_{2}-\Delta \mathrm{U}$
$=100-8.66-50$
$=41.34 \mathrm{~J}$
Similar Questions
A ball after falling from a height of $10\, m$ strikes the roof of a lift which is descending down with a velocity of $1\, m/s$. The recoil velocity of the ball will be .............. $\mathrm{m}/ \mathrm{s}$
A ball after falling from a height of $10\, m$ strikes the roof of a lift which is descending down with a velocity of $1\, m/s$. The recoil velocity of the ball will be .............. $\mathrm{m}/ \mathrm{s}$
An object has momentum $p$ & kinetic energy $E$. If its momentum becomes $2\,p$ then its kinetic energy will be :-
An object has momentum $p$ & kinetic energy $E$. If its momentum becomes $2\,p$ then its kinetic energy will be :-
A vertical spring with force constant $k$ is fixed on a table. A ball of mass $m$ at a height $h$ above the free upper end of the spring falls vertically on the spring so that the spring is compressed by a distance $d$. The net work done in the process is
A vertical spring with force constant $k$ is fixed on a table. A ball of mass $m$ at a height $h$ above the free upper end of the spring falls vertically on the spring so that the spring is compressed by a distance $d$. The net work done in the process is
A force of $\left( {2\widehat i + 3\widehat j + 4\widehat k} \right)\,N$ acts on a body for $4\, sec$ and produces a displacement of $\left( {3\widehat i + 4\widehat j + 5\widehat k} \right)\,m$. The power used is :- ............... $\mathrm{W}$
A force of $\left( {2\widehat i + 3\widehat j + 4\widehat k} \right)\,N$ acts on a body for $4\, sec$ and produces a displacement of $\left( {3\widehat i + 4\widehat j + 5\widehat k} \right)\,m$. The power used is :- ............... $\mathrm{W}$
Power supplied to a particle of mass $2\, kg$ varies with time as $P = \frac{{3{t^2}}}{2}$ $watt$ . Here, $t$ is in $seconds$ . If velocity of particle at $t = 0$ is $v = 0$, the velocity of particle at time $t = 2s$ will be ............. $\mathrm{m}/ \mathrm{s}$
Power supplied to a particle of mass $2\, kg$ varies with time as $P = \frac{{3{t^2}}}{2}$ $watt$ . Here, $t$ is in $seconds$ . If velocity of particle at $t = 0$ is $v = 0$, the velocity of particle at time $t = 2s$ will be ............. $\mathrm{m}/ \mathrm{s}$