a certain force acting on a body of mass 2 kg increases it's velocity from 6m/s to 15m/s in 2 seconds . the work done by the force during this interval is ?
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Sra
W = Fd
F = ma
a = (15-6)/2 = 4.5
d = 1/2 at^2 = 2.25*4 = 9
W = 2*4.5*9 = 81
To calculate the work done by a force, we need to use the formula:
Work = Force × Distance × cos(theta),
where:
- Work is the energy transferred by the force,
- Force is the magnitude of the force applied,
- Distance is the distance over which the force is applied,
- theta is the angle between the force and the direction of motion.
In this case, we need to find the force applied and the distance covered in order to calculate the work done.
First, let's find the force applied on the body. We know that force is equal to the rate of change of momentum, which can be calculated using the equation:
Force = (mass × change in velocity) / time.
Given:
- mass = 2 kg,
- initial velocity = 6 m/s,
- final velocity = 15 m/s,
- time = 2 seconds.
Substituting the values into the equation, we get:
Force = (2 kg × (15 m/s - 6 m/s)) / 2 s.
Simplifying further:
Force = (2 kg × 9 m/s) / 2 s
= 18 N.
Now that we have the force, we need to find the distance covered by the body. We can use the equation of motion:
Distance = (initial velocity × time) + (0.5 × acceleration × time^2),
where:
- initial velocity = 6 m/s,
- time = 2 seconds,
- acceleration = (final velocity - initial velocity) / time.
First, let's calculate the acceleration:
acceleration = (15 m/s - 6 m/s) / 2 s
= 9 m/s / 2 s
= 4.5 m/s^2.
Now, substituting the values into the equation:
Distance = (6 m/s × 2 s) + (0.5 × 4.5 m/s^2 × (2 s)^2)
= 12 m + (0.5 × 4.5 m/s^2 × 4 s^2)
= 12 m + (0.5 × 4.5 m/s^2 × 16 s^2)
= 12 m + 0.5 × 4.5 m/s^2 × 16 s^2
= 12 m + 0.5 × 4.5 m/s^2 × 16 s^2
= 12 m + 36 m
= 48 m.
Now that we have the force (18 N) and the distance (48 m), we can calculate the work done:
Work = Force × Distance × cos(theta),
Since the angle (theta) is not given, we assume the force is acting in the same direction as the displacement (zero angle), so the cos(theta) is equal to 1.
Substituting the values into the equation:
Work = 18 N × 48 m × 1
= 864 Nm.
Therefore, the work done by the force acting on the body during the interval is 864 Nm.