Question

A force of 3 N acts on a 6 kg cart. How long will it take before the cart speeds up to 4 m/s?

f= 3N
m=3kg

i need help with the formulas. do i first find the acceleration a = f/m? i don't know the next formula, how do i find how long it takes?

A cart goes 50 m in 5s. If the mass is 2.5 kg what force is needed?
m=2.5 kg
t=5s
i know that the force formula is f= m * a. theres no acceleration? how do we find it.

A 2 kg cart is pulled by an 8N force. How far will the cart have traveled when the speed hits 4m/s?
f=8n
m=2kg
we would use the distance formula. im not so sure what to do first

A force of 3N pulls a 2 kg cart down a hill. how much time will it take the cart to go down the hill if it is 3 m long?
f=3n
m=2kg
what formula is used here

A force of 10 N pulls a cart down a hill. The cart reaches a speed of 10 m/s after traveling a distance of 2 m. What is the mass of the cart?
do we just use the formula F/a = m? how do we find acceleration? is that the right formula im using

Answer 1

A force of 3 N acts on a 6 kg cart. How long will it take before the cart speeds up to 4 m/s?
f= 3N
m=6 kg NOTE not 3
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yes you have the right idea but write out for all your problems this:
F = m a
v = Vi + a t
x = Xi + Vi T + (1/2) a t^2
now
assume Vi = initial speed = 0
a = F/m = 3/6 = 0.5 m/s^2
v = 0 + a t
so v = 0 + 0.5 t
or 5 = 0 + .5 t
t = 10 seconds
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A cart goes 50 m in 5s. If the mass is 2.5 kg what force is needed?
m=2.5 kg
t=5s
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PERHAPS THE SPEED WAS ZERO at t = 0 (it should say if so)
I will assume that Vi = 0
v = 0 + a t where a = F/m = F/2.5
so
v = (F/2.5)(5)
NOW a trick to make it easier--- v versus t is straight line slope a
SO use average velocity for time and distance
Vav = 50/5 = 10 m/s
Vmax = 2 * Vav = 20 m/s
so
20 m/s = 0 + (F/2.5)(5 )
20 = 2 F
F = 10 Newtons
now check
a = 10 /2.5 = 4 m/s^2
v = a t = ( 4) (5) = 20 m/s at 5 seconds, right
V



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A 2 kg cart is pulled by an 8N force. How far will the cart have traveled when the speed hits 4m/s?
f=8n
m=2kg
we would use the distance formula. im not so sure what to do first
v = 0 + a t
4 = 0 + a t
but a = F//m = 8/2 = 4 m/s^2
so
4 = 0 + 4 t
t = 1 second
now
x = 0 + 0 t + (1/2)(4)(1^2) = 2 meters
tough huh ?
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A force of 3N pulls a 2 kg cart down a hill. how much time will it take the cart to go down the hill if it is 3 m long?
f=3n
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well I guess that is a net force including m g and friction maybe?
a = 3/2
v = (3/2)t
x = 3 = (1/2)(3/2) t^2
4 = t^2
t = 2 seconds

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A force of 10 N pulls a cart down a hill. The cart reaches a speed of 10 m/s after traveling a distance of 2 m. What is the mass of the cart?
do we just use the formula F/a = m? how do we find acceleration? is that the right formula im using

again I guess mg is included in the 10 Newtons
a = 10/m
v = a t = 10 t
so
t = v/10
at x = 2 m, v = 10
so t at 2 m = 10/10 = 1 second
x = (1/2) a t^2
2 = (1/2) a *1^2
a = 4
4 = 10/m so m = 10/4