Advertisements
Advertisements
प्रश्न
A 100 kg lock is started with a speed of 2.0 m s−1 on a long, rough belt kept fixed in a horizontal position. The coefficient of kinetic friction between the block and the belt is 0.20. (a) Calculate the change in the internal energy of the block-belt system as the block comes to a stop on the belt. (b) Consider the situation from a frame of reference moving at 2.0 m s−1 along the initial velocity of the block. As seen from this frame, the block is gently put on a moving belt and in due time the block starts moving with the belt at 2.0 m s−1. calculate the increase in the kinetic energy of the block as it stops slipping past the belt. (c) Find the work done in this frame by the external force holding the belt.
Advertisements
उत्तर
Here,
m = 100 kg
u = 2.0 m/s
v = 0
μk = 0.2
a) Internal energy of the belt-block system will decrease when the block will lose its KE in heat due to friction. Thus,
KE lost = `1/2m u^2-1/2mnu^2`
`1/2m(u^2-nu^2)`
`1/2xx100xx(2^2-0^2)`
= 200 J
b) Velocity of the frame is given by
uf = 2.0 m/s
u’ = u - uf = 2 – 2= 0
v’ = 0 – 2 = -2 m/s
KE lost = `1/2m u'^2-1/2mnu'^2`
`=1/2m(0^2-nu'^2)`
`=1/2xx100xx(0^2-2^2)`
= 200 J
c) Force of friction is given by
`f=mu_k^""R`
`rArrf=0.2xxmg=0.2xx100xx10=200N
`"Retardation"=f/m=200/100=2ms^-2`
Distance moved by the block will be as seen from the frame = s
`nu'^2-u'^2=2as`
`rArr2^2-0^2=2xx2s`
`rArrs=1"m"`
Work done by the force responsible for accelaration as seen from the frame = fs
= 200 x 1 = 200J
Work done by the belt to give it a final velocity of 2 m/s
`=1/2mnu'^2`
`=1/2xx100xx(2)^2`
= 200J
Total work done by external force as seen from the frame 200 + 200 = 400J
APPEARS IN
संबंधित प्रश्न
Explain why Two bodies at different temperatures T1 and T2, if brought in thermal contact, do not necessarily settle to the mean temperature (T1 + T2)/2.
In changing the state of a gas adiabatically from an equilibrium state A to another equilibrium state B, an amount of work equal to 22.3 J is done on the system. If the gas is taken from state A to B via a process in which the net heat absorbed by the system is 9.35 cal, how much is the net work done by the system in the latter case? (Take 1 cal = 4.19 J)
When a tyre bursts, the air coming out is cooler than the surrounding air. Explain.
An ideal gas goes from the state i to the state f as shown in figure. The work done by the gas during the process ______________ .
Consider two processes on a system as shown in figure.
The volumes in the initial states are the same in the two processes and the volumes in the final states are also the same. Let ∆W1 and ∆W2 be the work done by the system in the processes A and B respectively.
In a process on a system, the initial pressure and volume are equal to the final pressure and volume.
(a) The initial temperature must be equal to the final temperature.
(b) The initial internal energy must be equal to the final internal energy.
(c) The net heat given to the system in the process must be zero.
(d) The net work done by the system in the process must be zero.
A substance is taken through the process abc as shown in figure. If the internal energy of the substance increases by 5000 J and a heat of 2625 cal is given to the system, calculate the value of J.
A mixture of fuel and oxygen is burned in a constant-volume chamber surrounded by a water bath. It was noticed that the temperature of water is increased during the process. Treating the mixture of fuel and oxygen as the system,
- Has heat been transferred?
- Has work been done?
- What is the sign of ∆U?
Explain given cases related to energy transfer between the system and surrounding –
- energy transferred (Q) > 0
- energy transferred (Q) < 0
- energy transferred (Q) = 0
Explain the different ways through which the internal energy of the system can be changed.
8 m3 of a gas is heated at the pressure 105 N/m2 until its volume increases by 10%. Then, the external work done by the gas is ____________.
Two cylinders A and B of equal capacity are connected to each other via a stopcock. A contains a gas at standard temperature and pressure. B is completely evacuated. The entire system is thermally insulated. The stopcock is suddenly opened. Answer the following:
What is the final pressure of the gas in A and B?
Figure shows the P-V diagram of an ideal gas undergoing a change of state from A to B. Four different parts I, II, III and IV as shown in the figure may lead to the same change of state.
- Change in internal energy is same in IV and III cases, but not in I and II.
- Change in internal energy is same in all the four cases.
- Work done is maximum in case I
- Work done is minimum in case II.
n mole of a perfect gas undergoes a cyclic process ABCA (see figure) consisting of the following processes:
A `→` B: Isothermal expansion at temperature T so that the volume is doubled from V1 to V2 = 2V1 and pressure changes from P1 to P2.
B `→` C: Isobaric compression at pressure P2 to initial volume V1.
C `→` A: Isochoric change leading to change of pressure from P2 to P1.
Total workdone in the complete cycle ABCA is ______.
A cyclic process ABCA is shown in the V-T diagram. A process on the P-V diagram is ______.
The internal energy of one mole of argon is ______.
A steam engine delivers 4.8 x 108 Jof work per minute and services 1.2 x 109 J of heat per minute from its boiler. What is the percentage efficiency of the engine?
What is heat?
A system releases 125 kJ of heat while 104 kJ work is done on the system. Calculate the change in internal energy.
