Work Done on 100 kg Block by 100 N Force in 4s

x2017
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Homework Statement



A 100 N net force acts on a 100 kg block (initially at rest) for 4s
a. How much work was done on the block? (1)
b. What is the block’s velocity after the 4 s? (1)
c. How much energy does the block have after this time?(1)
d. The 100 N net force is no longer acting on the block. If the block slides an additional 3 m before coming to rest, what average force must have acted on the block over those 3 m? (1)

Homework Equations


P=W/t
work done = change in energy
Hooke's Law: Fs=kΔx
Ws=(1/2)kΔx2
KE=(1/2)mv2
PEg=mgh
SE=(1/2)kΔx2
Total energy = KE+PE+SE
W=FΔd

The Attempt at a Solution


The only information I have is:
F=100N
m=100kg
Δt=4s

All of the equations I have require information I don't have so I am very confused as to how to attempt this question.

W=FΔd was what I thought I should be using, but I don't have Δd. Then I thought maybe this one: Ws=(1/2)kΔx2, but I don't have k OR Δx... :frown:
 
on Phys.org
x2017 said:
A 100 N net force acts on a 100 kg block (initially at rest) for 4s
You can calculate acceleration from this. Use kinematical equations to find the displacement.
 
cnh1995 said:
You can calculate acceleration from this. Use kinematical equations to find the displacement.

The kinematic equation I have for acceleration is
a=(vf-vi)/Δt
a=(vf-0)/4

I don't have a final velocity though, just F, m and Δt.
x2017 said:
A 100 N net force acts on a 100 kg block (initially at rest) for 4s
 
x2017 said:
A 100 N net force
x2017 said:
100 kg block
Remember Mr. Newton..
 
x2017 said:
vf=Ft/m
vf=(100)(4)/100
vf=4m/s
That is true because initial velocity=0.
F=m(vf-vi/t.
In general, F=ma
You have m=100kg, F=100N.
This gives a=1m/s2...
Since
v=u+at, and u=0,
v=at=1*4=4m/s...
 

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