Question
A 2.0 kg object starts from rest from the top of an inclined plane that makes an angle of 30 degrees with the horizontal and has a rise of .4m. It slides down the inclined plane and at the bottom has a kinetic energy of 60J. Ignore friction and use g = 10.m/s2. what is the speed of the object at the bottom of the inclined plane?
Answers
Tsega
0.45
Shi yu
Evaporations is water turning into air and making clouds.
Anonymous
34
Helping
The energy at the bottom will be equal to Kinetic Energy. PE=0
Use KE =(1/2)mv^2
60 J = (1/2)(2 kg)v^2
60 J = v^2
v = 7.7 m/s
Use KE =(1/2)mv^2
60 J = (1/2)(2 kg)v^2
60 J = v^2
v = 7.7 m/s