Solve Normal Force: 1.4kg Ball Dropped, Find Average Force

[petern]

Problem:
A 1.4 kg ball is dropped from an unknown height, h, and hits the ground with a velocity of 15.0 m/s and rebounds with a velocity of 11.0m/s. If the ball is in contact with the ground for 1/35 of a second, find the average force the ground applies to the ball. (Use GUESS method.) Answer: 1288 N.

Given: m = 1.4 kg, V-initial = 15 m/s, V-final = 11.0 m/s, t = 1/35 s

Unknown: average Fn ground applies

E:
WE:

S:

S:

Can someone please get me started?

 

[PhanthomJay]

Try writing down your givens and unknowns and equations you think you might be using. If the ball hits at 15m/s down and returns at 11m/s up, what is its velocity change during the period it is in contact with the ground? what would be its average acceleration during that 1/35 second time period? Then think Newton 2, identifying all forces acting on the ball during this period.

 

[petern]

The force the ground applies must be calculated by multiplying the mass of the object by the object's acceleration, thus we must first find acceleration using kinematics equations and the given information
v2 = v1 + at
11.0m/s [Up] = -15m/s [Up] + a(1/35s)
26m/s [Up] = a(1/35s)
26m/s[Up]/(1/35s) = a
910m/s [Up] = a

now that we know acceleration we use Newton's second law (F = am) to calculate the force the ground is exerting on the ball

F = 910m/s [Up] X 1.4kg
F = 1274 N

The force that the ground applies to the ball is 1274N

How come the final answer doesn't match up with the given answer of 1288N?

 

[PhanthomJay]

The force the ground applies must be calculated by multiplying the mass of the object by the object's acceleration, thus we must first find acceleration using kinematics equations and the given information
v2 = v1 + at
11.0m/s [Up] = -15m/s [Up] + a(1/35s)
26m/s [Up] = a(1/35s)
26m/s[Up]/(1/35s) = a
910m/s [Up] = a

now that we know acceleration we use Newton's second law (F = am) to calculate the force the ground is exerting on the ball

F = 910m/s [Up] X 1.4kg
F = 1274 N

The force that the ground applies to the ball is 1274N

How come the final answer doesn't match up with the given answer of 1288N?

Yes, good up to a point. Newton's law says that the net force is ma . You therefore have calculated the net force. You are looking for the normal force. What other forces are acting on the ball?

 

[petern]

Yes, good up to a point. Newton's law says that the net force is ma . You therefore have calculated the net force. You are looking for the normal force. What other forces are acting on the ball?

I absolutely cannot figure it out. The only force I can think of is normal force and gravity. There's no friction because you're not given any data on it.

 

[PhanthomJay]

I absolutely cannot figure it out. The only force I can think of is normal force and gravity. There's no friction because you're not given any data on it.

Gravity, yes! What is the gravity force acting on the mass?? Which way does it act??

 

[petern]

Gravity, yes! What is the gravity force acting on the mass?? Which way does it act??

Yay! Down both when the ball is going down and when it is going up after it has rebounded.