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
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
For the first question, to find the time it takes for the cart to speed up to 4 m/s, you are on the right track with using the formula for acceleration.
First, find the acceleration using the formula a = F/m, where F is the force (3 N) and m is the mass of the cart (6 kg). So, a = 3 N / 6 kg = 0.5 m/s^2.
To find the time it takes for the cart to reach a speed of 4 m/s, you can use the formula v = u + at, where v is the final velocity (4 m/s), u is the initial velocity (0 m/s), a is the acceleration (0.5 m/s^2), and t is the time we want to find.
Rearrange the formula to solve for t: t = (v - u) / a. Plugging in the values, t = (4 m/s - 0 m/s) / 0.5 m/s^2 = 8 seconds. So it will take 8 seconds for the cart to speed up to 4 m/s.
For the second question, you are correct that the force formula is f = m * a. In this case, you are given the distance traveled and the time, so you can find the acceleration first.
Using the formula v = s/t, where v is the velocity (50 m/5 s = 10 m/s), s is the distance (50 m), and t is the time (5 s). The distance traveled is not related to acceleration, so you can't find the acceleration using this information.
Without the acceleration, you cannot find the force directly using the formula f = m * a. You would need more information to calculate the force required.
For the third question, to find how far the 2 kg cart will have traveled when it reaches a speed of 4 m/s, you will need to use the formula for distance.
The formula to find distance is s = ut + 0.5at^2, where s is the distance, u is the initial velocity (0 m/s), a is the acceleration, and t is the time.
You have the force (8 N) and the mass of the cart (2 kg), but you still need to find the acceleration. Once you have the acceleration, you can use the formula to find the distance traveled.
For the fourth question, to find the time it takes for the 2 kg cart to go down a 3 m hill with a force of 3 N, you will use the formula for distance.
Again, you have the force (3 N) and the mass of the cart (2 kg), but you still need to find the acceleration. Once you have the acceleration, you can use the formula to find the time taken.
For the fifth question, to find the mass of the cart, you need to use the formula f = m * a.
In this case, you are given the force (10 N) and the speed (10 m/s) after traveling a distance (2 m). To find the acceleration, you can use the formula v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity (0 m/s), a is the acceleration, and s is the distance.
Rearranging the formula to solve for a: a = (v^2 - u^2) / (2s). Plugging in the values, a = (10 m/s)^2 / (2 * 2 m) = 25 m/s^2.
Finally, use the formula f = m * a to find the mass (m) of the cart. Rearranging the formula to solve for m: m = f / a. Plugging in the values, m = 10 N / 25 m/s^2 = 0.4 kg. So the mass of the cart is 0.4 kg.