4 Work-Kinetic energy theorem Apply th an object launched from the level of the

Question

4 Work-Kinetic energy theorem Apply th an object launched from the level of the ground with an initial velocity IVol e Work-Kinetic energy theorem to the problem of parabolic motion consider an angle of 45 degrees above horizontal. Call v, the velocity of the object when a. What is the work done by gravity from the starting point (0,0) to the top of itial-velocity vol = 3 m/s 3 m/s at an angle of 45 degrees above hori it hits the ground. Set the wero of potential energy at the ground level. the trajectory? (be careful: the top of the trajectory is the point where the height of the object is maximum.) b. What is the work done by gravity from the top of the trajectory down to the ground? c. What is the total work done by the gravitational force on the object? d. Now that you know the value of the total work, use the work energy theorem to find the magnitude of the final velocity v.

Explanation / Answer

(a) Gravity works only on vertical component of velocity so horizontal component of projectile would remain same

vertical component will vary because of the gravity

Initial vertical velocity was Vi Sin theta = 3 Sin45 = 2.12 m/s

at the top of the projectile its vertical velocity become zero

Work done = change in kinetic energy

W = K.Ef - K.Ei

W = 1/2 mVf2 - 1/2mVi2

W = 0 - 1/2 m (2.12)2

W = - 2.25 m

It is negative because it was acting opposite to the direction of vertical motion

(b) Work done from the top to the ground

Final vertical velocity at the bottom of the trajectory would also be same at initial but in downward direction

And here initial velocity is at the top of the trajectory

W = 1/2 mVf2 - 1/2mVi2

W = 1/2m*(2.12)2 - 0

W = 2.25 m

(c) Total work done = -2.25 +2.25 =0

(d) As the net work done is zero so final velocity would be same as initial

Vf = Vi = 3 m/s

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