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Question: \(300\;{\rm{J}}\) of work is done in sliding a \(2\;{\rm{kg}}\) block up an inclined plane of height...

300  J300\;{\rm{J}} of work is done in sliding a 2  kg2\;{\rm{kg}} block up an inclined plane of height 10  m10\;{\rm{m}}. Work done against friction is (Take g=10  m/s2g = 10\;{\rm{m}}/{{\rm{s}}^2})
(a) 1000  J1000\;{\rm{J}}
(b) 200  J200\;{\rm{J}}
(c) 100  J100\;{\rm{J}}
(d) zero

Explanation

Solution

The above problem is based on the work done under the effect of friction. The work done depends on the change in the position and direction of the force applied on the object. The loss of work occurs due to the presence of the friction.

Complete step by step answer
Given: The work done to slide the block is W=300  JW = 300\;{\rm{J}}, mass of the block is m=2  kgm = 2\;{\rm{kg}}, height of the is h=10  mh = 10\;{\rm{m}}, gravitational acceleration is g=10  m/s2g = 10\;{\rm{m}}/{{\rm{s}}^2}.
The formula to calculate the work done to raise the block on the inclined plane is given as:
W1=mgh{W_1} = mgh
Substitute 2  kg2\;{\rm{kg}}for m, 10  m/s210\;{\rm{m}}/{{\rm{s}}^2}for g and 10  m10\;{\rm{m}}for h in the above formula to find the work done to raise the block on the inclined plane.
W1=(2  kg)(10  m/s2)(10  m){W_1} = \left( {2\;{\rm{kg}}} \right)\left( {10\;{\rm{m}}/{{\rm{s}}^2}} \right)\left( {10\;{\rm{m}}} \right)
W1=200  J{W_1} = 200\;{\rm{J}}
The formula to calculate the work done against friction is given as:
Wf=W−W1{W_f} = W - {W_1}
Substitute 300  J300\;{\rm{J}}for WW and 200  J200\;{\rm{J}}for W1{W_1} in the above expression to find the work done against friction.
Wf=300  J−200  J{W_f} = 300\;{\rm{J}} - 200\;{\rm{J}}
Wf=100  J{W_f} = 100\;{\rm{J}}

Thus, the work done against friction by the block is 100  J100\;{\rm{J}} and the option (c) is the correct answer.

Additional Information: The work done by the object is divided into two types. First one is the work done by the conservative force and second one is the non-conservative force. The total mechanical energy of the system remains conserved if conservative forces are involved in the system. The change in the total energy remains conserved if non- conservative forces are involved in the system.

Note: The work done by the object is equal to the work done by the object against gravity and work done by the object against the friction. The work done by the object also equals the change in the kinetic energy of the object.