Since F=ma, the heavier cart has 1/3 the acceleration, so its velocity will be 1/3 that of the smaller cart.
KE = 1/2 mv^2, so if you have 1/3 v instead of v, the KE will be 1/9 as big.
So, K=20J for the larger cart.
Two carts of masses m and 3m are at rest on a horizontal frictionless track. A person pushes each cart with the same force F for 5 s. If the kinetic energy of the lighter cart after the push is K = 180J, the kinetic energy of the heavier cart is :
3 answers
I agree on the velocity of 1/3, but the bigger car has triple the mass, so we have KE=3*1/9=1/3 KE of the smaller, or 60J
force8time=momentum1=momentum2
momentum1=mometum2
m*v=3m (v/3)
KE1=1/2 mv^2=180J
KE2=1/2 3m(v/3)^2=1/2 mv^2/3
but = KE1/3= 60J
force8time=momentum1=momentum2
momentum1=mometum2
m*v=3m (v/3)
KE1=1/2 mv^2=180J
KE2=1/2 3m(v/3)^2=1/2 mv^2/3
but = KE1/3= 60J
Damon is, as usual right in these matters. I forgot to factor in the added mass.