- #1
physicslady123
- 10
- 0
- Homework Statement
- The ball with a mass of 200g is shot up with the instantaneous velocity of 14 m/s.
a) Determine the mechanical energy at the point of release
b) Find the speed of the ball after it has travelled for 0.2s
c) What is the kinetic energy of the ball after 0.2s
d) Use the conservation of energy principle to determine the max height the ball reaches
- Relevant Equations
- E(gravity)=mgh
E(kinetic)=1/2mv^2
E(mech1)=E(g)+E(k)
E(mech1)=E(mech2)
a) E(mech)=E(k)+E(g)
E(mech)=1/2mv^2+(0.2)(9.8)(0)
E(mech)=19.6 J
b) E(mech1)=E(mech2)
E(k)+E(g)=E(k)+E(g)
E(g)=E(k)+E(g)
0=1/2mv^2+(mgh)
*No height is given so I can't solve using this method. It says instantaneous velocity meaning the velocity at 0.2s is different.
c) E(k)=1/2mv^2
E(k)=1/2(0.2)(i would use the velocity from q.b)^2
d) E(mech1)=E(mech2)
E(k)+E(g)=E(k)+E(g)
E(g)=E(k)
mgh=1/2mv^2
h=v^2/2g
h=(how would i calculate v?)^2/2(9.8)
So, I am stuck on how to calculate a few values. I can't use kinematics as this is the energy unit.
E(mech)=1/2mv^2+(0.2)(9.8)(0)
E(mech)=19.6 J
b) E(mech1)=E(mech2)
E(k)+E(g)=E(k)+E(g)
E(g)=E(k)+E(g)
0=1/2mv^2+(mgh)
*No height is given so I can't solve using this method. It says instantaneous velocity meaning the velocity at 0.2s is different.
c) E(k)=1/2mv^2
E(k)=1/2(0.2)(i would use the velocity from q.b)^2
d) E(mech1)=E(mech2)
E(k)+E(g)=E(k)+E(g)
E(g)=E(k)
mgh=1/2mv^2
h=v^2/2g
h=(how would i calculate v?)^2/2(9.8)
So, I am stuck on how to calculate a few values. I can't use kinematics as this is the energy unit.
