a 1.5-kilogram cart initially moves at 2.0 meters per second. it is brought to rest by a constant net force in 0.30 second. what is the magnitude of the net force
Since acceleration is constant, we can use kinematic equation v=u+at( v= final velocity = 0...since finally cart stops. u = initial velocity = 2m/s, a = deacceleration and t= 0.3 sec). Therefore 0 = 2 - a(0.3). a = 20/3 = 6.67m/s^2. Net force = massxacceleration = 1.5x6.67 = 10N.
The magnitude of the net force is 10 N
From Newton's second law of motion, we have that
F = ma
Where F is the force
m is the mass
and a is the acceleration
Acceleration, a, is given by the formula
\(a = \frac{\Delta v}{ t}\)
Δv is the change in velocity
and t is the time
From the given information
Initial velocity of the cart is 2.0 m/s
and
Since the cart was brought to rest, that means the final velocity of the cart is 0 m/s
∴Δv = 2.0 m/s
and t = 0.30 secs
Putting the parameters into the equation, we get
\(a = \frac{2.0}{0.3}\)
∴ \(a = 6.67\ m/s^{2}\)
From the given information,
Mass of the cart = 1.5 kg
∴ Net force, F = 1.5 × 6.67
F = 10 N
Hence, the magnitude of the net force is 10 N
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