A 10-kg box Initially at rest and moves along a frictionless horizontal surface. A Horizontal force to the right as applied to the box. The magnitude of the box change of the function of time as shown
b. What is the velocity of the box at 10 seconds? Show your work.
c. If you want to apply at constant force starting at 10 seconds, that would stop the box at exactly 20 seconds. What would the magnitude and direction of that force be? Show your work
Answers
Answer:
b. 1.1 m/s c. -1.1 m/s
Explanation:
b.
m x v=f x t
The mass is 10 kg
10 x v (equation for momentum) = f x t (impulse)
you would find the impulse by finding the area under the graph, and then dividing by the mass
So in this case:
10 x v = 11
then divide by the mass
v = 11/10
v= 1.1 m/s
c.
F x t = impulse
we are trying to find force (magnitude)
impulse/t = F
we said that in order for it to stop the impulse for b would be 11, so if thats the case it would be 11 but it would be negative because when going in a positive direction and suddenly stopping the direction changes.
-11/t = F
to find t all we need to do is find the amount of seconds between 10 and 20, that would be 10.
-11/10 = F
F= -1.1 m/s
Answer B:
The velocity of the box at 10 seconds is :
Formula :m x v=f x t
Given Information :
Mass(m)=10kg
10 x v (equation for momentum) = f x t (impulse)10 x v = 11v = 11/10v= 1.1 m/sTherefore, the velocity of the box at 10 seconds is 1.1m/s.
Answer C:
The magnitude and direction of that force be :
Formula:F x t = impulse
impulse/t = F-11/t = FTo find t all we need to do is find the amount of seconds between 10 and 20, that would be 10.
-11/10 = FF= -1.1 m/sThe magnitude and direction of that force will be -1.1m/s.
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