Calculate the braking force to bring a body of mass 1kg to rest from 25m/s on a level ground in 60m with uniform retardation?
V^2 = Vo^2 + 2a*d = 0,
25^2 + 2a*60 = 0,
a = -5.2 m/s^2.
F = M*a = 1 * (-5.2) = -___N.
The negative sign means the force opposes the motion.
To calculate the braking force needed to bring a body of mass 1kg to rest from 25m/s on a level ground in 60m with uniform retardation, we can use the equations of motion.
We can start by determining the deceleration (retardation) of the body. We know that the initial velocity (u) is 25m/s, the final velocity (v) is 0m/s (since the body comes to rest), and the displacement (s) is 60m.
We can use the equation: v^2 = u^2 + 2as, where:
- v is the final velocity (0 in this case),
- u is the initial velocity (25m/s),
- a is the deceleration (which we need to find),
- s is the displacement (60m).
Rearranging the equation, we get: a = (v^2 - u^2) / (2s)
Substituting the given values, we have: a = (0 - 25^2) / (2 * 60)
Simplifying the equation, we get: a = -625 / 120 = -5.21 m/s²
The negative sign indicates that the body is decelerating or slowing down.
Next, we can calculate the braking force (F) using the equation: F = ma, where:
- m is the mass of the body (1kg in this case),
- a is the deceleration (-5.21 m/s²).
Substituting the values, we have: F = 1kg * (-5.21 m/s²)
Simplifying the equation, we get: F = -5.21 N
Therefore, the braking force required to bring a body of mass 1kg to rest from 25m/s on a level ground in 60m with uniform retardation is approximately -5.21 Newtons (N). The negative sign indicates that the force acts in the opposite direction to the motion.
To calculate the braking force needed to bring a body to rest, we can use Newton's second law of motion, which states that force is equal to mass times acceleration (F = ma). In this case, the acceleration is the uniform retardation.
First, let's find the acceleration using the given information. We know the initial velocity (u) is 25 m/s, the final velocity (v) is 0 m/s (since the body comes to rest), and the distance traveled (s) is 60 m.
We can use the following equation to find the acceleration (a):
v^2 = u^2 + 2as
Rearranging the equation, we have:
a = (v^2 - u^2) / (2s)
Plugging in the values:
a = (0^2 - 25^2) / (2 * 60)
a = (-625) / 120
a ≈ -5.21 m/s²
Since retardation is a negative acceleration, we assign it a negative sign. Therefore, the retardation is approximately 5.21 m/s².
Now, we can calculate the braking force (F) using the formula F = ma. The mass (m) is given as 1 kg, and the acceleration (a) is approximately 5.21 m/s².
F = 1 kg * 5.21 m/s²
F ≈ 5.21 N
Hence, the braking force required to bring the body to rest from 25 m/s in 60 m with uniform retardation is approximately 5.21 Newtons.
If it takes t seconds, then we have
25-at = 0
25t - 1/2 at^2 = 60
Solve that for a, and then as usual, F=ma