Question

A car weighing 1400 kg is moving at a speed of 54 km/h up a hill when the motor stops. If it is just able to reach the destination which is at the height of 10 m above the point, Calculate the work done against friction (Negative of the work done by the friction).
Concept of Physics - 1 , HC VERMA , Chapter " Work and Energy"

Answer 1

Here given in the question :-
Car Mass m = 1400 kg .
h = 10 m
Moving at Initial speed v= 54 km/h = (54000/3600) m/s = 15 m/sec.

Now Initial Kinetic Energy of the car ,
= 1/2 mv²
= 1/2 (1400 x 15²)
= 315000/2
= 157500 J.

Here final velocity and Final Kinetic energy both are Zero.
Now change in Kinetic Energy 
= Final - Initial
= 0 - 157500
= -157500 J.
So change in Kinetic energy is also equal to W.D by all force.

Thus work done by gravity,
= mgh
= 1400 x 9.8 (-10)
= 1400 x -98
= -137200 .

Here, change in Kinetic energy = Work done by the friction + Work done by the gravity.
-157500 = Work done by friction + (- 137200)
Work done by the friction = -157500 +137200
$W _{f}$ =  -20,300

Hence, the work done against the friction =  20300 J


Hope it Helps. :-)

Answer 2

Answer:

To reach the minimum time he has to move with maximum possible acceleration.

Let the maximum acceleration be a.

ma-μR=0

ma=μmg

a=μg=0.9x10=9 m/s2

A) Initial velocity=u= o m/s

t=?

a=9m/s2  

s=5m

From s= ut +1/2 at²

50=0+1/2 at²

t=100/9

t=10/3 sec=3.3 sec

b) AFTER MOVING 50m, velocity of the athelete given by :

V=u+at=0+9x 10/3

he has to stop in minimum time, so deceleration is  

-a=-9 m/s2  

R=mg

ma=μR

a=μg =9m/s2[deceleration]

u1=30m/s

v1=0 m/s]t=v1-u1/a

t=0-30/-a

t=-30/-9

t=10/3 sec=3.3 sec