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 :
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.
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
Damon is, as usual right in these matters. I forgot to factor in the added mass.
To find the kinetic energy of the heavier cart, we can use the principle of conservation of energy. According to this principle, the total energy of a closed system remains constant.
At the beginning, both carts are at rest, so their total kinetic energy is zero. After the push, the kinetic energy of the lighter cart is given as K = 180 J.
We can start by calculating the force applied to the lighter cart. Since the carts have the same force F applied to them, we can use Newton's second law, which states that the force applied to an object is equal to the mass of the object multiplied by its acceleration (F = ma).
For the lighter cart, the force F applied for 5 seconds results in an acceleration a. Using the equation F = ma, we have F = m * a. Rearranging the equation, we get a = F/m.
Since the carts are on a horizontal frictionless track, the force applied creates an acceleration in the forward direction. Therefore, the net force is equal to the force applied.
Now, we can calculate the acceleration of the lighter cart using the given force F and mass m.
a = F/m
To calculate the kinetic energy of the heavier cart, we need to know how the force F affects the acceleration of the lighter cart. If the force F applied is the same for both carts, the larger mass of the heavier cart will result in a smaller acceleration.
Let's assume that the acceleration of the heavier cart is a'. Since the force applied to the heavier cart is the same as the force applied to the lighter cart, we can write F = 3m * a'.
Now, we will use the concept of work.
The work done on an object is the force applied to it multiplied by the distance it moves (W = F * d). In this case, the distance traveled by both carts is the same.
The work done on the lighter cart, which is equal to the change in kinetic energy, is given as K = 180 J.
Similarly, the work done on the heavier cart, which is equal to the change in kinetic energy, can be written as W' = F * d = 3m * a' * d, where d is the distance traveled by both carts.
Since the work done is equal to the change in kinetic energy, we can write:
W' = 3m * a' * d = K
Now, we can divide the equation for the lighter cart acceleration (a = F/m) by the equation for the heavier cart acceleration (a' = F/(3m)) to eliminate the force:
a / a' = (F/m) / (F/(3m))
Simplifying, we get:
a / a' = 3
This means that the acceleration of the lighter cart is three times greater than the acceleration of the heavier cart.
Since the distance traveled by both carts is the same, we can relate their respective kinetic energies:
K / K' = (1/2) * m * a^2 / (1/2) * 3m * a'^2
Simplifying, we get:
K / K' = a^2 / (3a'^2)
Substituting the acceleration relationship found above:
K / K' = a^2 / (3(a/3)^2) = 1 / 1/3 = 3
The ratio of the kinetic energy of the lighter cart to the kinetic energy of the heavier cart is 3.
Therefore, the kinetic energy of the heavier cart is:
K' = K / ratio = 180 J / 3 = 60 J
Hence, the kinetic energy of the heavier cart is 60 J.