Work-energy theorem: It states that the sum of work done by all the forces acting on a body is equal to the change in the kinetic energy of the body i.e.,
The work done by a person in carrying a box of mass 15 kg through a vertical height of 5 m is 5000 J. The mass of the person is _______ Take g = 10 m/s^{2}
95 kg
45 kg
65 kg
85 kg
Answer (Detailed Solution Below)
Option 4 : 85 kg
Work Power and Energy MCQ Question 3 Detailed Solution
A drum of mass ‘m’ kg was rolled down a ramp. At the bottom of the ramp its kinetic energy was 10 kJ and velocity was 20 m/s. With what velocity (in m/s) was it pushed down the ramp if its initial kinetic energy was 625 J?
5
7.5
10
2.5
Answer (Detailed Solution Below)
Option 1 : 5
Work Power and Energy MCQ Question 7 Detailed Solution
Kinetic energy (KE): The energy possessed by a body by virtue of its motion is called kinetic energy.
\(KE = \frac{1}{2}m{v^2}\)
Where m = mass of the body and v = velocity of the body
Potential energy (PE): The energy possessed by a body by virtue of its position or configuration is called potential energy.
\(PE = mgh\)
Where, m = mass of the body, g = acceleration due to gravity and h = height of the body
Conservation of energy: According to the conservation of energy, energy can’t be created or destroyed, it can only transformfrom one form to another.
The total energy of the system remains constant i.e. the sum of kinetic energy + potential energy remains constant.
EXPLANATION:
When the body will be thrown upward the initial kinetic energy will slowly get converted into potential energy. but the total energy will remain constant.
Thus the potential energy will increase and the kinetic energy will decrease until the body reached the top. So option 2 is correct.
An object of mass 8 kg is moving with a momentum of 16 kg m/s. A force of 0.4 N is applied to it in the direction of the motion of the object for 20 sec. The increase in the kinetic energy is ________.
20 J
40 J
10 J
32 J
Answer (Detailed Solution Below)
Option 1 : 20 J
Work Power and Energy MCQ Question 10 Detailed Solution
Kinetic energy (KE): The energy possessed by a body by virtue of its motion is called kinetic energy.
\(KE = \frac{1}{2}m{v^2}\)
Where m = mass of the body and v = velocity of the body
Momentum (P): The product of mass and velocity is called momentum. The SI unit of momentum is kg m/s.
Momentum (P) = Mass (m) × Velocity (v)
Force (F): The interaction which after applying on a body changes or try to change the state of motion or state of rest is called force.
Force (F) = Mass (m) × acceleration (a)
The equation of motion is given by:
v = u + a t
Where v is final velocity, u is initial velocity, a is acceleration and t is time.
CALCULATION:
Given that:
⇒ Mass of the object (m) = 8 kg , Momentum (P) = 16 kg m/s, Applied force (F) = 0.4 N, Time (t) = 20 sec
⇒ We know that, Momentum (P) = Mass (m) × velocity (v)
⇒ 16 = 8 × u
⇒ u = 2 m/s = Initial velocity
⇒ Force (F) = Mass (m) × Acceleration (a)
0.4 = 8 × a
⇒ Acceleration (a) = 0.05 m/s^{2}
⇒ Final velocity (v) = u + at = 2 + 0.05 × 20 = 3 m/s
⇒ So, change in kinetic energy = Final KE - Initial KE =\(\frac{1}{2}mv^2 -\frac{1}{2}mu^2=\frac{1}{2}.8.[v^2 -u^2]=\frac{1}{2}.8.[3^2 -2^2]=\frac{1}{2}.8.[9 -4]=20 \; J\)
⇒ Thus, theincrease in the kinetic energy is 20 J.
Work done (W): When a force is applied on a body and there is a displacement of the body in the direction of force then the work is said to be done by the force.
Work: When a force is applied on a body and there is a displacement of the body in the direction of force then the work is said to be done by the force.
Find the power possessed by an object of mass 5 kg when it is at a height of 4 m above the ground and it takes 2 seconds to reach the given height slowly. (take, g = 10 m s^{–2})
200 W
100 W
50 W
400 W
Answer (Detailed Solution Below)
Option 2 : 100 W
Work Power and Energy MCQ Question 13 Detailed Solution
Work is said to be done when a force applied on an object causes displacement of the object.
Work done (W) = force (F) × displacement (S)
Work-energy theorem: It states that the sum of work done by all the forces acting on a body is equal to the change in the kinetic energy of the body i.e.,
A bicyclist comes to a skidding stop in 10 m. During this process, the force on the bicycle due to the road is 200 N and is directly opposed to the motion. The work done by the cycle on the road is:
Zero
-2000 J
1 J
2000 J
Answer (Detailed Solution Below)
Option 1 : Zero
Work Power and Energy MCQ Question 16 Detailed Solution
Work is said to be done by a force when the body is displaced actually through some distance in the direction of the applied force.
Since the body is being displaced in the direction of F, therefore work done by the force in displacing the body through a distance s is given by:
\(W = \vec F \cdot \vec s\)
Or, W = Fs cos θ
Thus work done by a force is equal to the scalar or dot product of the force and the displacement of the body.
CALCULATION:
The Cycle has applied brakes. The tyre is now skidding. The person must have put some force on the pedals that the bike stops only after 10 m and not immediately(sudden stop).
Given that:
The Displacement D, of the cycle, is 10 m
The Frictional Force acting on the cycle is 200 N.
The Force the cycle applies will be equal to frictional force = 200 N.
The road does not move, D or displacement of the road is zero,