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Question: A body of mass 2 kg is being dragged with a uniform velocity of 2 m/s on a rough horizontal plane. T...

A body of mass 2 kg is being dragged with a uniform velocity of 2 m/s on a rough horizontal plane. The coefficient of friction between the body and the surface is 0.20, J = 4.2 J/Cal and g = 9.8 m/s29.8\text{ m/}{{\text{s}}^{2}}. Calculate the amount of heat generated in 5 sec.
A. 9.33 Cal
B. 19.33 Cal
C. 29.33 Cal
D. 39.33 Cal

Explanation

Solution

As the body is dragged, a frictional force acts on the body. The frictional force acting on the body is calculated using the coefficient of friction and the normal reaction force acting on the body. The work done on the body is then calculated using the force of friction acting on the body and distance (calculated as a product of speed and time) through which the body is moved. Work done (calculated in joules) is then divided by J = 4.2 J/Cal, which gives us the heat energy in calories.
Formula used:
Distance is given by,
d=v×td=v\times t
Where, v is velocity and t is time.
The force of reaction acting on the body is given by,
R=mgR=mg
Where, m is mass and g is acceleration due to gravity.
Frictional force acting on the body is given by,
F=μRF=\mu R
Where, μ\mu is coefficient of friction and R is normal force acting on the body.
Work done is given by,
W=F×dW=F\times d
Where, F is force and d is distance.
Heat energy is given by,
Q=WJQ=\dfrac{W}{J}
Where, W is work done and J is mechanical equivalent of heat.

Complete step by step answer:
The distance through which the body is dragged with a velocity of 2 m/s for 5 seconds is calculated as:
d=v×td=v\times t
 =2×5\text{ }=2\times 5
 =10 m\text{ }=10\text{ m}
Force acting on the body is calculated as:
R=mgR=mg
 =2×9.8\text{ }=2\times 9.8
 =19.6 N\text{ }=19.6\text{ N}
Frictional force is calculated as:
F=μRF=\mu R
 =0.20×19.6\text{ }=0.20\times 19.6
 =3.92 N\text{ }=3.92\text{ N}
Work done is calculated as:
W=F×SW=F\times S
 =3.92×10\text{ }=3.92\times 10
 =39.2 J\text{ }=39.2\text{ J}
Heat energy produced by the work done is calculated as:
Q=WJQ=\dfrac{W}{J}
 =39.2 J4.2 J/cal\text{ }=\dfrac{39.2\text{ J}}{4.2\text{ J/cal}}
 =9.33 cal\text{ }=9.33\text{ cal}

The answer is option A.

Note:
The force acting on the body can be calculated directly by multiplying the coefficient of friction, mass of the body and acceleration due to gravity, that is, F=μmgF=\mu mg. This is obtained by combining the formulas for normal force and frictional force.