please include the steps
To warm up for a match, a tennis player hits the 57.0 g ball vertically with her racket.
If the ball is stationary just before it is hit and goes 5.20 m high, what impulse did she impart to it?
the answer should be in kg*m/s
mass : m = 57.0 gr = 5.70 x 10^-2 kg
acceleration of gravity : g = 9.8 m/s²
maximum high : h = 5.20 m
● speed of the ball (after it was beaten) :
=> v ' = √[2 g h]
=> v ' = √[2 (9.8) (5.20)] = 10.096 m/s
● Impulse = change in momentum
=> I = ∆p
=> I = m v ' - m v . . . . . . . (initially are still → v = 0)
=> I = m v '
=> I = (5.70 x 10^-2) (10.096)
=> I = 0.575 kg m/s
Indeed that is *not* correct. In order to do work a force has to act over a distance. So you would have to lift the dumbbell not just hold it. It may be true that the respiration reaction in you muscles (C6H12O6 (s) + 6 O2 (g) → 6 CO2 (g) + 6 H2O (l) + heat) needs to produce energy at a rate of 1 hp over and above the normal rate in order to hold the weight. But I wouldn't know about the exact quantities. Oh, I see you already realise that. Well it seems to me that muscles are always 'working' anyway in the sense that they are constantly converting sugar into CO2, H20 an heat so it may be that this reaction produces enough energy to account for the 1 hp. Just to note for the scenario of a person climbing a ladder that there will be work done by the muscles lifting the person against gravity, all of that comes from respiration, not all of the respiration energy will be used by the muscles to lift the person however. That's ok Jon. Just came across your question and thought I'd give it a go.:)
If the ball goes up to a height of 5.20 m, then its initial velocity (velocity as it leaves the racket) is
V = sqrt (2gh)
g= acceleration due to gravity = 9.8 m/sec^2 (constant)
h = 5.20 m (given)
V = sqrt(2*9.8*5.2)
V = 10.09 m/sec.
Impulse = change in momentum = (57/1000)(10.09 - 0)
Impulse = 5.75 kg-m/sec.
Hope this helps.
impulse is change in momentum.
momentum is mV = (0.57)(V)
but first we have to calculate V, which we can get from potential and kinetic energy, setting them equal and solving for V
Kinetic Energy in J
KE = ½mV²
Potential Energy in J
PE = mgh
mgh = ½mV²
gh = ½V²
9.8*5.2 = ½V²
V = 10.1 m/s
back to the momentum equation
P = mV = (0.57)(10.1) = 5.75 kgm/s