Solveeit Logo
Login

Question

Question: What is work done by a person in carrying a suitcase weighing \(10\,kg\) on his head when he travels...

What is work done by a person in carrying a suitcase weighing 10 kg10\,kg on his head when he travels a distance of 5m5m in the (i) vertical direction and (ii) horizontal direction? Take g=9.8 msec⁡−2g = 9.8\,m{\sec ^{ - 2}}

Explanation

Solution

In order to solve this question we need to understand work done and force. Work done is defined as the energy which is stored in an object when a force acts on it causing it to displace some distance. It is mathematically expressed as a scalar product of force and displacement. In this question we would try to find the angle between force and displacement in both cases. Also work can be positive, negative and zero depending on the angle between force and displacement.

Complete step by step answer:
Mass of an object is given as, M=10 kgM = 10\,kg
Acceleration of object is given as, g=9.8 msec⁡−2g = 9.8\,m{\sec ^{ - 2}}
For part (a), Let the vertical direction be positive and denoted as j^\hat j
So displacement in this case is, d⃗1=5j^{\vec d_1} = 5\hat j
Since the force of gravity is given as, F⃗=−Mgj^\vec F = - Mg{\kern 1pt} \hat j
Here, negative sign indicates that the force of gravity is in downward direction,
So putting values we get, F⃗=−(10)(9.8)j^\vec F = - (10)(9.8){\kern 1pt} \hat j
F⃗=−98j^\vec F = - 98{\kern 1pt} \hat j
So work is defined as, W=F⃗.d⃗1W = \vec F.{\vec d_1}
Putting values we get, W=(−98j^).(5j^)W = ( - 98{\kern 1pt} \hat j).(5\hat j)
W=−490J(j^.j^)W = - 490J(\hat j.\hat j)
∴W=−490J\therefore W = - 490J Since we know, j^.j^=1\hat j.\hat j = 1

So here work is done, W=−490 JW = - 490\,J.

For part (b), Let the horizontal direction be positive and denoted as i^\hat i
So displacement in this case is, d⃗2=5i^{\vec d_2} = 5\hat i
Since the force of gravity is given as, F⃗=−Mgj^\vec F = - Mg{\kern 1pt} \hat j
Here, negative sign indicates that the force of gravity is in downward direction,
So putting values we get, F⃗=−(10)(9.8)j^\vec F = - (10)(9.8){\kern 1pt} \hat j
F⃗=−98j^\vec F = - 98{\kern 1pt} \hat j
So work is defined as, W=F⃗.d⃗2W = \vec F.{\vec d_2}
Putting values we get, W=(−98j^).(5i^)W = ( - 98{\kern 1pt} \hat j).(5\hat i)
W=−490J(j^.i^)W = - 490J(\hat j.\hat i)
∴W=0 J\therefore W = 0\,J Since we know, j^.i^=0\hat j.\hat i = 0

So here work is done, W=0 JW = 0\,J.

Note: It should be remembered that according to the work energy theorem, work done by force is equal to change in kinetic energy of the body. Also work is conservative in nature if the force is conservative in nature. Force is conservative when the potential or potential difference depends only on space and does not depend on time.