Average force exerted on a ball?

[IAmPat]

Homework Statement

A tennis player strikes a tennis ball of mass 56.7g when it is at the top of the toss, accelerating it to 68.0m/s in a distance of 0.0250 m. What is the average force the player exerts on the ball? Ignore any other forces acting on the ball.

Homework Equations

Net-Force = mass * acceleration
Acceleration = Change in velocity / Change in time

The Attempt at a Solution

Change in velocity = 68
Change in time = 0.0250 seconds

Acceleration = 68/0.025 = 2,720 m/s/s
mass = 56.7g -> 0.0567kg

Net-Force = (0.0567)(2,720) = 154.224 N



I'm able to check if my answer is correct after submitting it so I know whether or not it is correct. I thought this would be the only way to do it, but perhaps I am not finding the average force? Any help is appreciated

 

[gomunkul51]

You were given: "a distance of 0.0250 m"
and you used it as: "Change in time = 0.0250 seconds"
this is obviously incorrect.

You can use the Work-Energy Theorem to solve this problem !

 

[IAmPat]

You were given: "a distance of 0.0250 m"
and you used it as: "Change in time = 0.0250 seconds"
this is obviously incorrect.

You can use the Work-Energy Theorem to solve this problem !

Wow, all this time and I didn't catch that error. Thanks for pointing that out.

As for the work-energy theorem, would I just be using this?

Work = .5 * m * Vf
Work = .5 * 0.0567kg * 68m/s
Work = 131.09 N

Or am I missing a step somewhere?

 

[SammyS]

Work done by the net force = change in Kinetic Energy.

W = F_NET·x

ΔKE = (½)m(v_f²) - (½)m(v₀²)

 

[IAmPat]

Work done by the net force = change in Kinetic Energy.

W = F_NET·x

ΔKE = (½)m(v_f²) - (½)m(v₀²)

I had a mistake in writing out my last post.

Work = .5 * m * Vf
Work = .5 * 0.0567kg * 68^2 m/s
Work = 131.09 N

Is in fact right, mathematically atleast.

ΔKE = (½)m(vf2) - (½)m(v02)
ΔKE = (1/2)(0.0567)(68^2) - (1/2)(0.0567)(0^2)
ΔKE = 131.09 N

I get the same answer. The initial velocity is 0 (right?), because it's just being tossed straight up in the air. Unfortunately 131.09N is still wrong apparently. I don't know what I'm doing wrong.

 

[SammyS]

I had a mistake in writing out my last post.

Work = .5 * m * Vf² This is the final KE.
Work = .5 * 0.0567kg * 68^2 m/s
Work = 131.09 N·m
Work is force times distance. The racket applies a force on the ball over a distance of 0.250m.

Is in fact right, mathematically at least.

ΔKE = (½)m(vf2) - (½)m(v02)
ΔKE = (1/2)(0.0567)(68^2) - (1/2)(0.0567)(0^2)
ΔKE = 131.09 N·m  

I get the same answer. The initial velocity is 0 (right?), because it's just being tossed straight up in the air. Unfortunately 131.09N is still wrong apparently. I don't know what I'm doing wrong.

See comments in red above.

 

[IAmPat]

See comments in red above.

Ah, thank you.

I divided 131.09 by distance and got

131.09/0.0250 = 5243.6 N

Which was correct. Thanks

 

[gomunkul51]

yup.