Work, Energy and Simple Machines — CBSE Class 9 Science Important Questions

Answer: (c) Joule.

Work $=$= force $\times$× displacement, so its unit is $\text{N}\times\text{m}=\text{N m}=\text{joule}$N×m=N m=joule. Energy, being the capacity to do work, has the same unit.

Answer: (c) Zero.

Gravity acts vertically downward while the displacement is horizontal, so the angle between force and displacement is $90^\circ$90^. Since $W=Fs\cos90^\circ=0$W=Fs90^=0, the work done against gravity is zero.

Answer: (c) Four times.

Kinetic energy $E_k=\tfrac12mv^2\propto v^2$E_k=12mv^2 v^2. Doubling $v$v multiplies $v^2$v^2 by $4$4, so the kinetic energy becomes four times its original value.

Answer: (b) $3.6\times10^6\ \text{J}$3.6×10^6 J.

$1\ \text{kWh}=1000\ \text{W}\times3600\ \text{s}=3.6\times10^6\ \text{J}$1 kWh=1000 W×3600 s=3.6×10^6 J. It is the commercial unit of electrical energy.

Answer: (a) Both A and R are true and R is the correct explanation of A. Since the centripetal force is directed toward the centre, perpendicular to the velocity, $\theta=90^\circ$=90^ and $W=Fs\cos90^\circ=0$W=Fs90^=0.

Power is the rate of doing work (or of transferring energy): $P=\dfrac{W}{t}$P=W/t.

Its SI unit is the watt ($\text{W}$W). One watt is the power of an agent that does $1\ \text{joule}$1 joule of work in $1\ \text{second}$1 second: $1\ \text{W}=1\ \text{J s}^{-1}$1 W=1 J s^-1.

Work: $W=Fs=15\times3=45\ \text{J}$W=Fs=15×3=45 J.

Power: $P=\dfrac{W}{t}=\dfrac{45}{5}=9\ \text{W}$P=W/t=45/5=9 W.

Consider a body of mass $m$m at rest ($u=0$u=0). A constant force $F$F acts on it over a displacement $s$s, giving it acceleration $a$a and final velocity $v$v.

Work done $=Fs=(ma)s$=Fs=(ma)s.

Using $v^2=u^2+2as$v^2=u^2+2as with $u=0$u=0: $v^2=2as$v^2=2as, so $as=\dfrac{v^2}{2}$as=v^2/2.

Therefore the work done, which is stored as kinetic energy:
$E_k=m(as)=m\cdot\dfrac{v^2}{2}=\tfrac12mv^2.$E_k=m(as)=m·v^2/2=12mv^2.

Potential energy at the top:
$E_p=mgh=10\times10\times5=500\ \text{J}.$E_p=mgh=10×10×5=500 J.

Kinetic energy just before landing: By the law of conservation of energy, all the potential energy converts to kinetic energy (ignoring air resistance):
$E_k=E_p=500\ \text{J}.$E_k=E_p=500 J.

Power $=60\ \text{W}=0.06\ \text{kW}$=60 W=0.06 kW.

Daily usage $=0.06\ \text{kW}\times6\ \text{h}=0.36\ \text{kWh}$=0.06 kW×6 h=0.36 kWh.

Energy in $30$30 days:
$E=0.36\times30=10.8\ \text{kWh}.$E=0.36×30=10.8 kWh.

Thus $10.8$10.8 units of electrical energy are consumed.

Law of conservation of energy: energy can neither be created nor destroyed; it only changes from one form to another, and the total energy of an isolated system remains constant.

Free fall (taking $g$g as acceleration, ignoring air resistance):

At the top (height $H$H, at rest): $E_p=mgH$E_p=mgH, $E_k=0$E_k=0, so total $=mgH$=mgH.

At a point after falling a distance $x$x: velocity $v^2=2gx$v^2=2gx, height $=(H-x)$=(H-x).
$E_k=\tfrac12mv^2=\tfrac12m(2gx)=mgx,\quad E_p=mg(H-x).$E_k=12mv^2=12m(2gx)=mgx, E_p=mg(H-x).
$\text{Total}=mgx+mg(H-x)=mgH.$Total=mgx+mg(H-x)=mgH.

Just before landing ($h=0$h=0): $v^2=2gH$v^2=2gH, so $E_k=\tfrac12m(2gH)=mgH$E_k=12m(2gH)=mgH, $E_p=0$E_p=0, total $=mgH$=mgH.

At every stage the total mechanical energy equals $mgH$mgH; the potential energy steadily converts into kinetic energy, confirming conservation of energy.

(a) A simple machine is a device that helps us do work more conveniently by changing the magnitude or direction of an applied force, e.g. a lever, pulley or inclined plane.

Mechanical advantage (MA) is the ratio of the load lifted to the effort applied:
$\text{MA}=\dfrac{\text{load}}{\text{effort}}.$MA=load/effort.

(b) $\text{MA}=\dfrac{200}{50}=4$MA=200/50=4. Since MA $>1$>1, the lever multiplies the effort four times.

(c) In a class-I lever the fulcrum lies between the load and the effort:

(i) At the top the stored water possesses gravitational potential energy.

(ii) Potential energy $\rightarrow$→ kinetic energy (of falling water) $\rightarrow$→ mechanical energy (spinning turbine) $\rightarrow$→ electrical energy (generator).

(iii) $E_p=mgh=500\times10\times80=4\times10^{5}\ \text{J}=400000\ \text{J}$E_p=mgh=500×10×80=4×10^5 J=400000 J.

(iv) By conservation of energy with no losses, all potential energy converts to kinetic energy:
$E_k=E_p=4\times10^{5}\ \text{J}=400000\ \text{J}.$E_k=E_p=4×10^5 J=400000 J.