A horizontal cable pulls a 200 kg cart along a horizontal track.the tension in the cable is 500N. Starting from rest (a)how long will it take the cart to reach a speed of 8m/s (b) how far will it have gone ?
To solve this problem, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration. We'll also use the kinematic equation, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken.
Given:
Mass of the cart, m = 200 kg
Tension in the cable, T = 500 N
Initial velocity, u = 0 m/s
Final velocity, v = 8 m/s
(a) Finding the acceleration:
Using Newton's second law, we can determine the acceleration experienced by the cart:
T - frictional force = ma
The frictional force is usually negligible, so:
T = ma
rearranging the equation, we get:
a = T/m
a = 500 N / 200 kg = 2.5 m/s^2
Now, using the kinematic equation, we can find the time taken to reach a speed of 8 m/s:
v = u + at
8 m/s = 0 + 2.5 m/s^2 * t
8 = 2.5t
t = 8 / 2.5 = 3.2 s
Therefore, it will take the cart 3.2 seconds to reach a speed of 8 m/s.
(b) Finding the distance traveled:
To find the distance traveled by the cart, we can use the kinematic equation:
s = ut + (1/2)at^2
As the cart starts from rest, the initial velocity u is 0, so the equation simplifies to:
s = (1/2)at^2
substituting the values:
s = (1/2) * 2.5 m/s^2 * (3.2 s)^2
s = 6.4 m
Therefore, the cart will have gone a distance of 6.4 meters.
To answer those questions, we can use the equations of motion to calculate the time taken and the distance traveled by the cart.
(a) To find the time it takes for the cart to reach a speed of 8 m/s, we'll use the equation:
v = u + at
Where:
v = final velocity (8 m/s)
u = initial velocity (0 m/s, since the cart starts from rest)
a = acceleration
First, we need to find the acceleration by using Newton's second law:
F = ma
Where:
F = tension in the cable (500 N)
m = mass of the cart (200 kg)
a = acceleration (unknown)
Solving for acceleration:
a = F/m
a = 500 N / 200 kg
a = 2.5 m/s^2
Now, we can use the equation of motion to find the time:
v = u + at
8 m/s = 0 m/s + 2.5 m/s^2 * t
Rearranging the equation for t:
t = (v - u) / a
t = (8 m/s - 0 m/s) / 2.5 m/s^2
t = 3.2 seconds
Therefore, it will take the cart 3.2 seconds to reach a speed of 8 m/s.
(b) To find the distance the cart will have gone, we can use the equation:
s = ut + (1/2)at^2
Where:
s = distance
u = initial velocity (0 m/s)
t = time (3.2 seconds)
a = acceleration (2.5 m/s^2)
Plugging in the values:
s = 0 m/s * 3.2 seconds + (1/2) * 2.5 m/s^2 * (3.2 seconds)^2
Simplifying the equation:
s = 0 + (1/2) * 2.5 m/s^2 * 10.24 seconds^2
s = 12.8 meters
Therefore, the cart will have traveled a distance of 12.8 meters.
F = ma
v = at
s = 1/2 at^2