What is the kinetic energy of the object at t= 1.0 seconds?

Question

At t= 1.0 seconds, a 0.83 kg  object is falling with a speed of 6.0 m/s . At t=2.0 seconds, it has a kinetic energy of 25 J .

b.) What is the speed of the object at t=2.0 seconds?
c.) How much work was done on the object between t=1.0 seconds  and  t=2.0 seconds?

Can someone work this our for me step by step so that I can better understand it? Thanks!

Hola, muchacho · Accepted Answer

KE = 1/2mv^2
25 = (1/2)0.83v^2
sqrt(50/0.83) = v = 7.76 m/s


the object has fallen an additional second.
at this point, you can use kinematic formulas to find the distance after 1 second while setting initial speed = 7.76
d = 7.76*1 + (1/2)g(1)^2
g = 9.81
plug g in, you get: 12.665.
now the force acting on this ball is mg( the weight. )
0.86*9.81 = 8.44
12.665*8.44 = 106.9 our work yay (force times distance)

What is the kinetic energy of the object at t= 1.0 seconds?

1 Antwort