Work Energy & Power
Q1. The force between a pair of protons is repulsive. Does the PE increases or decreases as the protons are brought closer?
Ans. PE will increase as work has to be done against the force of repulsion. This is stored in the form of PE. PE decrease when a proton and an electron are brought nearer. Work will be done by the force of attraction between them.
Q2. Can mechanical energy of a body be negative? Justify.
Ans. Yes mechanical energy can be negative. Mechanical energy is the sum of potential as well as kinetic energy.
Negative sign in the energy does not have anything to do with the direction. Hence, being scalar has no bearing on the fact that energy cannot be negative.
The kinetic energy is always positive, but the potential energy can be negative.
Negative sign only indicates that the system is either bound or the reference point is taken such that potential energy is negative.
Q3. A light body and a heavy body both have the same kinetic energy. Which one will have the greater momentum?
Ans. KE = 1/2 x m x v x v = (mv x mv)/2m = p x p/ 2m or p x p = 2mKE or p x p = m
Thus heaviour body has a greater momentum than the lighter one.
Q4. Define an electron –volt. Express it in joule.
Ans. 1 electron volt is the energy change that takes place when a charge equal to 1 electron (1.6×10-19 C) is moved through a potential difference of 1 volt.
1eV = 1.602×10-19 J
Q5. Throwing mud on wall is an example of perfectly inelastic collision. Explain.
Ans. When mud is thrown on a wall, it sticks to the wall. The kinetic energy of the mud is reduced to zero and there is non-conservation of kinetic energy. Hence it is a case of perfectly inelastic collision.
Q6. A gardener pushes a lawn roller through a distance of 20m. If he applies a force of 20kg wt in a direction inclined at 600 to the ground, find the work done by him.
Ans. F = 20 Kg-wt = 20 x 9.8 N
θ = 60 and s = 20 m
Work = Fs Cos θ = 20 x 9.8 x 20 x Cos 60 = 1960 J
Q7. A person is holding a bucket by applying a force of 10N. He moves a horizontal distance of 5 m and then climbs up a vertical distance of 10m. Calculate the total work done by him.
Ans. Let θ be the angle between force and displacement vectors. For horizontal motion,
F=10 N, s=5 m, θ=90°
Work done,W=→F.→s
Work done,
W1=Fscosθ=10×5×cos90°=0
For vertical motion, the angle between force and displacement is 0°.
Here,F=10 N, s= 10 m, θ=0°
Work done,W2=10×10×cos(0)=100 J
Total work done =W1+W2=100 J
Q8. Find the work doing in moving a particle along a vector s = (4i – j +7k ) m if the applied force is F = (i+2j+k) ?
Ans. Work done is defined as the displacement of the particle in the direction of applied force.
W = FS = (i + 2j - k) (4-j +7k) = 1x4 +2(-1) + (-1)x7 = 4-2-7 = -5 Nm
Therefore, the work done in moving a particle is -5N.m or 5J.
Q9. A particle is acted upon by force = (2) N and = () N, is displace from the point A(2, 1, 0) to the point B(-3, -4, 2). Find the total work done by these forces.
Ans. F1 = 2i-3j+4k and F2 = -i +2j -3k and Fnet = F1 + F2 = i-j+k
rA = 2i +j and rB = -3i-4j+2k and S=rB-rA = (-3i-4j+2k) - (2i + j) = -5i -5j + 2k
Wnet = Fnet . S = (i-j+k).(-5i-5j+2k) = -5 +5 +2 = 2J
Q10. A force F= a +bx acts on a particle in the x-direction, where a and b are constants. Find the work done by this force during a displacement from x= 0 to x=d.
Q11. A body of mass 4 kg initially at rest is subjected to a force of 16N. What is the kinetic energy acquired at the end of 10s?
Ans. m = 4kg , u = 0, F = 16 N, t = 10s, F=ma
16=4a, a = 4ms-2
v=u+at = 0 + 4x10
v = 40 ms-1
KE = 1/2 mv2
KE = 1/2 x 4 x (40) x (40) = 3200 J
Q12. If the linear momentum of a body increases by 20%, what will be the % increase in the kinetic energy of the body?
Q13. If the kinetic energy of a body increases by 300%, by what will the linear momentum of the body increase?
Q14. Calculate the velocity of the bob of a simple pendulum at its mean position if it is able to rise to a vertical height of 10cm.
Q15. A ball of mass m is dropped from a height h on a platform fixed at the top of a vertical spring. The platform is depressed by a distance x. What is the spring constant k?
Q16. A ball at rest is dropped from a height of 12 m. If it loses 25% of its KE in striking the ground, find the height to which it bounces. How do you account for the loss in kinetic energy?
Q17. The potential energy of a spring when stretched through a distance is x is 10 J. What is the amount of work done on the same spring to stretch it through an additional distance x?
Q18. A ball moving with the speed of 9m/s strikes an identical ball at rest, such that after collision, the direction of each ball makes an angle of 30°with the original line of motion. Find the speed of the two balls after collision.
Q19. An electric motor is used to lift an elevator of 1000kg and its load of 500kg to a height of 20m. The time taken for the job is 20s. What is the work done? What is the rate at which work is done? If the efficiency of motor is 75%, at which rate is the energy supplied to the motor?
Q20. A machine gun fires 60 bullets per minute with a velocity of 700m/s. If each bullet has a mass of 50g, find the power developed by the gun.
Ans. Each bullet will have KE = 1/2 x 0.05 x 700 x 700 = 12250 J
So for 60 bullets the energy given by machine in 60 second = 60 x 12250 J
Hence power developed = Energy / time = 12250 W
Or power = 12.250 kW.
Q21. A well 20 m deep and 3 m deep in diameter contains water to a depth of 14m. How long will a 5hp engine take to empty it? (1hp = 746W)
Q22. A man cycles up a hill whose slope is 1 in 20 with a velocity of 6.4 km/h. The weight of the man and the cycle is 98 kg. What work done per minute is he doing? What is his horse power?
Q23. A 10kg ball and 20kg ball approach each other with velocity 20m/s and 10m/s respectively. What are their velocities after collision if the collision is perfectly elastic?
Q25. A ball is dropped to the ground from a height of 2m. The coefficient of restitution is 0.6. to what height will the ball rebound?
Q24. A railway carriage of mass 9000kg moving with a speed of 36 km/h. What is their common speed after collision? What type of collision is it?
Q26. A steel ball of mass m moving with a speed of 10m/s collides with an identical ball at rest. After collision first ball moves at an angle of 30° with the original direction on one side whereas the second ball starts moving at an angle of 45° in the opposite side of original direction. Find the speed of each ball after the collision.