vertical component m/s (downward) horizontal component m/s (to the right) A part
ID: 1284341 • Letter: V
Question
vertical component m/s (downward) horizontal component m/s (to the right) A particle of mass 0.450 kg is shot from P as shown in Figure P7.6. The particle has an initial velocity vector V0. m/s upward (b) Determine the work done by the gravitational force on the particle during its motion from P to B. (Assume that h2 = 50.0 m.) J (c) Determine the horizontal and the vertical components of the velocity vector when the particle reaches B. vertical component m/s (downward) horizontal component m/s (to the right) Figure P7.6 (a) Using the law of conservation of energy, determine the vertical component of vector V 0 with a horizontal component of 30.0 m/s. The particle rises to a maximum height of h = 22.0 m above P.Explanation / Answer
a.) For the vertical component, you know the vertical velocity (final) = 0...
Vf*2 - Vi*2 = 2ay
0 - Vi*2 = 2(9.8 m/s*2)(22 m)
Vi(vertical) = 20.765 m/s
b.) Work (done by gravity) = Fd = mad =0.45 kg(9.8 m/s*2)(50 m)
W = 220.5 J
When the mass goes up the work is negative, so you only have to consider the distance from P down to B...
Notice that that's also its change in potential energy...
c.) The horizontal component is still 30 m/s...
The vertical component is...
Vf*2 - Vi*2 = 2ay
Vf*2 - 0 = 2(9.8 m/s*2)(50 + 22 m)
Vf = 37.566 m/s
I did this from the masses highest point where y velocity is 0...
This is not the only way to do this part...you can also use conservation of energy...
At the top of the flight KE(vertical) is 0 and at the ground PE is zero...
so mgh = 1/2mv*2
and 2gh = v*2
v*2 = 2(9.8)(50+22)
v = 37.566m/s
Hope that helped...
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