Find the velocity for a 20.0 gram bullet with the kinetic energy of 2000 Joules.

Question

Find the velocity for a 20.0 gram bullet with the kinetic energy of 2000 Joules.

A 2.0 kg rock is dropped from a height of 10.0 m above the earth's surface. Complete the table below by finding the values for the rock's potential, kinetic and total energy for the positions listed as it falls. Ignore air resistance.

Height (m) Potential Energy (J) Kinetic Energy (J) Total Energy (J)

10.0 ______________ ________________ ________________
5.0 _______________ _________________ ________________
1.0 _______________ __________________ __________________

Explanation / Answer

Given Mass of the bullet , m = 20 *10^-3 kg Kinetic energy of the bullet is , KE = 2000 J Since , KE = 1/2 mv^2       2000 J = 0.5 *(20 *10^-3 kg) *v^2           v = 447.21 m/s The velocity of the bullet is 447.21 m/s _________________________________________ Mass of the rock , m = 2.0 kg Height the rock dropped is , h = 10 m Potential energy at h1 = 10 m is , PE = mgh1 = 2.0kg*9.8 m/s^2*10 m = 196 J Kinetic energy at h1 is, KE =0 J Total energy is 196 J ------------------------------------------------------------------------- PE at h2 =5 m is , mgh2 = 2.0kg*9.8 m/s^2*5 m = 98 J KE at h2 = 5 m is mg(h1-h2) = KE KE = mg (5m) KE = 98 J Total energy , is 98J +98J = 196 J -------------------------------------------------- PE at h3 = 1 m is , mgh3 = 2.0kg*9.8 m/s^2*1m =19.6 J KE at h3 =1 m is mg(h1-h3) = KE KE = 2.0kg*9.8 m/s^2*9m  = 176.4 J Total energy , 19.6 J+ 176.4 J = 196 J KE = 98 J mg(h1-h3) = KE KE = 2.0kg*9.8 m/s^2*9m  = 176.4 J Total energy , 19.6 J+ 176.4 J = 196 J KE = 98 J

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