5. Solve the following examples.
a. An electric pump has 2 kW power.
How much water will the pump lift
every minute to a height of 10 m?
(Ans : 1224.5 kg)
b. If a 1200 W electric iron is used
daily for 30 minutes, how much
total electricity is consumed in the
month of April? (Ans:18 Unit)
c. If the energy of a ball falling from a
height of 10 metres is reduced by
40%, how high will it rebound?
(Ans: 6 m)
d. The velocity of a car increase from
54 km/hr to 72 km/hr. How much is
the work done if the mass of the car
is 1500 kg ?
(Ans. : 131250 J)
e. Ravi applied a force of 10 N and
moved a book 30 cm in the direction
of the force. How much was the
work done by Ravi?
(Ans: 3 J)
Answers
Answer:
a. Power , P=2 kW
Power , P=
TimeTaken(T)
Workdone(W)
⇒W=P×T
For every minute i.e. 60 s, work done by the pump is W=2×1000×60=120000J
Now, this work done is stored in water as it potential energy. Thus, mgh = 120000 J
⇒m=
9.8×10
120000
=1224.5 kg
Hence, 1224.5 kg of water is lifted by the pump every minute to a height of 10 m.
b. total energy from 10m height is
mgh = m*g*10 = 10mg
and when it is reduced by 40%
then amount of energy left = (100-40)% = 0.6*(10mg ) = 6mg
now mgh' = 6mg
so, h' = 6m
c. Given,
Initial height of ball, h=10m
Potential energy=mgh
Reduce potential energy = mgh
new
mgh
new
=(
100
100−40
)mgh
h
new
=0.6×10=6m
Hence, height after rebound is 6m.
d. Initial Velocity, u=54km/hr=15m/s
Final velocity , v=72km/hr=20m/s
Mass =1500kg
Initial kinetic energy
2
1
mu
2
⇒
2
1
1500×15×15=168750J
Final kinetic energy
2
1
mv
2
⇒
2
1
1500×20×20=300000J
So, change in kinetic energy = work done=300000−168750=131250J.
e. Work done ,
W=F×S×cosθ
Here, θ=0
∘
F=10NS=30cm=0.03m
∴W=10×0.03×1=3J.
Answer:
A) In the image
B) Power =1200W=1.2kWh
Time = 30 min / day × 30 days =0.5h×30= 15 h
Energy =Power×Time=1.2×15= 18kWh.
C) Initial height of ball, h=10m
Potential energy=mgh
Reduce potential energy = mghnew
mghnew=(100100−40)mgh
hnew=0.6×10=6m
Hence, height after rebound is 6m
D) Initial Velocity, u=54km/hr=15m/s
Final velocity , v=72km/hr=20m/s
Mass =1500kg
Initial kinetic energy 21mu2
⇒211500×15×15=168750J
Final kinetic energy 21mv2
⇒211500×20×20=300000J
So, change in kinetic energy = work done=300000−168750=131250J
E) Work done ,
W=F×S×cosθ
Here, θ=0∘
F=10NS=30cm=0.03m
∴W=10×0.03×1=3J
Explanation: