1. What is the speed of a 0.150 kg baseball if its kinetic energy is 110 J ? Answer in units of m/s
2. Two 4.7 g bullets are fired with speeds of 41.8 m/s and 53.8 m/s, respectively.
a) What is the kinetic energy of the first bullet? Answer in units of J
b) What is the kinetic energy of the second bullet? Answer in units of J
c) c) What is the ratio K2/K1 of their kinetic energies
3. Sally’s speed changes to 11.2 m/s. Now what is her kinetic energy? Answer in units of J
4. Tim, with mass 53.8 kg, climbs a gymnasium rope a distance of 5.9 m.
The acceleration of gravity is 9.8 m/s2 . How much potential energy does Tim gain? Answer in units of J
2. Two 4.7 g bullets are fired with speeds of 41.8 m/s and 53.8 m/s, respectively.
a) What is the kinetic energy of the first bullet? Answer in units of J
b) What is the kinetic energy of the second bullet? Answer in units of J
c) c) What is the ratio K2/K1 of their kinetic energies
3. Sally’s speed changes to 11.2 m/s. Now what is her kinetic energy? Answer in units of J
4. Tim, with mass 53.8 kg, climbs a gymnasium rope a distance of 5.9 m.
The acceleration of gravity is 9.8 m/s2 . How much potential energy does Tim gain? Answer in units of J
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Woah! The answers to these questions seem as simple as it is to read. Anyways
1) Using Kinetic energy formula 1/2mv^2, We se that
110J = 1/2* 0.150 * v^2. Therefore, v = [(110*2)/0.150]^1/2 = 38.3 m/s
2) a) K1 = 1/2mv^2 = 1/2*0.0047*41.8*41.8 = 4.1J
b)K2 = 1/2mv^2 = 1/2*0.0047*53.8*53.8 = 6.8J
c) K2/K1 = 1.65
3) Insufficient data
4) PE de to gravity = mgh = 53.8*9.8*5.9 = 3110.7 J
Hope this helps
1) Using Kinetic energy formula 1/2mv^2, We se that
110J = 1/2* 0.150 * v^2. Therefore, v = [(110*2)/0.150]^1/2 = 38.3 m/s
2) a) K1 = 1/2mv^2 = 1/2*0.0047*41.8*41.8 = 4.1J
b)K2 = 1/2mv^2 = 1/2*0.0047*53.8*53.8 = 6.8J
c) K2/K1 = 1.65
3) Insufficient data
4) PE de to gravity = mgh = 53.8*9.8*5.9 = 3110.7 J
Hope this helps