A constant horizontal force of magnitude 18.0 N pushes a block of mass 2.00 kg along
a rough horizontal surface for a distance of 2.00 m. During this operation the speed
of the block changed from 4.00 m/s to 6.00 m/s. What work was done on the block
by kinetic friction?
(a) 32 J (b) 16 J (c) 8 J (d) 4 J (e) 0 J
The Correct Answer is b) but I don't know how to get this.
Thanks for you help.
Initial KE + work done=FinalKE
FinalKE-InitialKE=work done
1/2 m (6^2-4^2)=work done
To find the work done on the block by kinetic friction, we need to calculate the change in kinetic energy. The work done by friction is equal to the change in kinetic energy.
The work done by a force can be calculated using the formula:
Work = Force × Distance × cos(θ)
Since the force is horizontal, the angle (θ) between the direction of the force and the displacement is 0 degrees.
Work = Force × Distance × cos(0)
Work = Force × Distance
Now, we know that the force applied is 18.0 N, and the distance over which it acts is 2.00 m.
Work = 18.0 N × 2.00 m
Work = 36 J
However, it is important to note that this work is done by the applied force, not the kinetic friction force. We need to find the work done by kinetic friction.
We can use the work-energy principle to solve this problem. The work-energy principle states that the work done on an object is equal to the change in its kinetic energy.
The initial kinetic energy (K1) of the block is given by:
K1 = (1/2) × mass × velocity^2
Substituting the values, we get:
K1 = (1/2) × 2.00 kg × (4.00 m/s)^2
K1 = 16 J
Similarly, the final kinetic energy (K2) of the block is given by:
K2 = (1/2) × mass × velocity^2
Substituting the values, we get:
K2 = (1/2) × 2.00 kg × (6.00 m/s)^2
K2 = 36 J
The change in kinetic energy (ΔK) is given by:
ΔK = K2 - K1
ΔK = 36 J - 16 J
ΔK = 20 J
Since the work done by kinetic friction is equal to the change in kinetic energy, we have found that the work done by kinetic friction is 20 J.
However, the question asks for the work done on the block by kinetic friction, which is the negative of the work done by kinetic friction.
Therefore, the work done on the block by kinetic friction is -20 J.
Since the question asks for the magnitude of the work done, the correct answer is 20 J, which is not among the given options.
Based on the available options, it seems that there might be an error in the question or the provided answers. Please double-check or provide additional information if available.