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

Respuesta :

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

[tex]a = \frac{\Delta v}{ t}[/tex]

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

[tex]a = \frac{2.0}{0.3}[/tex]

∴ [tex]a = 6.67\ m/s^{2}[/tex]

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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