A force of 5.0 N acts on a 15 kg body initially at rest. Compute the work done by the force in (a) the first,(b) the second, and (c) the third seconds and (d) the instantaneous power due to the force at the end of the third second.
To solve this problem, we need to use the following formulas:
1. Work (W) = force (F) × displacement (d) × cos(θ)
2. Power (P) = work (W) / time (t)
Given:
Force (F) = 5.0 N
Mass (m) = 15 kg
(a) To find the work done in the first second:
Here, the body is initially at rest, so the displacement is 0.
Using the formula W = F × d × cos(θ), and since cos(θ) = cos(0) = 1, we can calculate the work done as:
W1 = F × 0 × cos(0) = 0
Therefore, the work done in the first second is 0.
(b) To find the work done in the second second:
Here, the body is still at rest in the first second, so the displacement is also 0.
Using the formula W = F × d × cos(θ), and since cos(θ) = cos(0) = 1, we can calculate the work done as:
W2 = F × 0 × cos(0) = 0
Therefore, the work done in the second second is 0.
(c) To find the work done in the third second:
Here, the body is finally in motion. The force acting on the body is 5.0 N, and we need to calculate the displacement for one second of motion. To do this, we can use the formula:
F = m × a, where a is the acceleration.
Since the force is constant, we can rearrange the formula to find the acceleration:
a = F / m = 5.0 N / 15 kg = 0.33 m/s^2
Now, we can use the formula for displacement with constant acceleration:
d = (1/2) × a × t^2
= (1/2) × 0.33 m/s^2 × (3 s)^2
= 0.99 m
Using the formula W = F × d × cos(θ), and since cos(θ) = cos(0) = 1, we can calculate the work done as:
W3 = F × 0.99 m × cos(0) = 5.0 N × 0.99 m × 1 = 4.95 J
Therefore, the work done in the third second is 4.95 J.
(d) To find the instantaneous power at the end of the third second:
Using the formula P = W / t, we can calculate the power as:
P = 4.95 J / 3 s = 1.65 W
Therefore, the instantaneous power at the end of the third second is 1.65 W.
To compute the work done by a force, you need to multiply the force by the distance the object moves in the direction of the force.
In this scenario, the force acting on the body is 5.0 N. However, we don't have information about the distance the body moves. Therefore, we cannot directly compute the work done by the force.
Would you like to provide any additional information or ask any other questions related to the problem?
I'm guessing we're ignoring friction.
a = F/m
v = at
and KE = Work = 1/2mv^2
for part d) P = W/t