A constant force of friction of 50 N is acting on a body of mass 200kg moving initially with a speed of 15m/s. How long does the body take to stop? What distance will it cover before coming to rest?
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To find the time it takes for the body to stop, we can use the equation:
Acceleration = Force / Mass
Since the force of friction is acting in the opposite direction of the object's motion, the acceleration will be negative.
Acceleration = -50 N / 200 kg
Acceleration = -0.25 m/s^2
We can use the equation of motion to find the time it takes for the body to stop:
v = u + at
Where:
v = final velocity (0 m/s, since the body comes to rest)
u = initial velocity (15 m/s)
a = acceleration (-0.25 m/s^2)
t = time taken to stop
0 = 15 m/s + (-0.25 m/s^2) * t
Simplifying the equation:
0 - 15 m/s = -0.25 m/s^2 * t
-15 m/s = -0.25 m/s^2 * t
Dividing both sides by -0.25 m/s^2:
t = -15 m/s / -0.25 m/s^2
t = 60 seconds
Therefore, it will take 60 seconds for the body to come to a stop.
To find the distance it will cover before coming to rest, we can use the equation:
Distance = Initial Velocity * Time + (1/2) * Acceleration * Time^2
Since the final velocity is 0, the equation simplifies to:
Distance = Initial Velocity * Time
Distance = 15 m/s * 60 s
Distance = 900 meters
Therefore, the body will cover 900 meters before coming to rest.
To find out how long it takes for the body to stop, we can use Newton's second law of motion, which states that the net force acting on an object is equal to its mass multiplied by its acceleration. In this case, the net force acting on the body is the frictional force, and the acceleration is the rate at which the body is decelerating.
First, let's calculate the acceleration of the body using Newton's second law. The friction force is given as 50 N, and the mass of the body is 200 kg. Therefore,
Acceleration (a) = Net Force / Mass
a = 50 N / 200 kg
a = 0.25 m/s²
Now that we have the acceleration, we can use kinematic equations to find out how long it takes for the body to stop and the distance it covers.
1. To find the time it takes for the body to stop, we can use the equation:
Final velocity (v) = Initial velocity (u) + acceleration (a) × time (t)
Since the final velocity is 0 (the body comes to rest), the equation becomes:
0 = 15 m/s + (-0.25 m/s²) × t
Solving for time, we get:
t = -15 m/s / -0.25 m/s²
t = 60 s
So, the body takes 60 seconds to stop.
2. To find the distance it covers before coming to rest, we can use another kinematic equation:
Distance (s) = (Initial velocity (u) × time (t)) + (0.5 × acceleration (a) × time (t)²)
Plugging in the values:
s = (15 m/s × 60 s) + (0.5 × -0.25 m/s² × (60 s)²)
s = 900 m + (-450 m)
s = 450 m
Therefore, the body will cover a distance of 450 meters before coming to rest.
So, in summary:
- The body takes 60 seconds to stop.
- The distance covered before stopping is 450 meters.
M*g = 200*9.8 = 1960 N. = Wt. of the body. = Normal(Fn).
Fk = u*Fn = 50 N.
u * 1960 = 50.
u = 0.0255.
a = u*g = 0.0255 * (-9.8) = -0.25 m/s^2.
V = Vo + a*t = 0.
15 -0.25t = 0.
0.25t = 15.
t = 60 s.
d = Vo*t + 0.5a*t^2.
d = 15*t - 0.125*60^2 = 450 m.