A 1200-kg ore cart is rolling at 10.8 m/s across a flat friction-free surface. a crane suddenly drops of ore vertically into the
1 answer:
The value of the final speed depends on the mass of the ore.
Let's call m the mass of the ore. We can solve the exercise by requiring the conservation of momentum, which must be the same before and after the ore is loaded.
Initially, there is only the cart, so the momentum is
After the ore is loaded, the new mass will be (1200 kg+m), and the new speed is
. The momentum p is conserved, so it is still 12960 kg m/s. Therefore, we have
and so the final speed is
Let's call m the mass of the ore. We can solve the exercise by requiring the conservation of momentum, which must be the same before and after the ore is loaded.
Initially, there is only the cart, so the momentum is
After the ore is loaded, the new mass will be (1200 kg+m), and the new speed is
and so the final speed is
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Answer:
Friction between the box and the floor is 25N to the left
Explanation:
There are two forces acting on the box along the horizontal direction:
- The force of push applied by the worker, in the forward direction, F
- The force of friction, , acting in the opposite direction (backward)
So the net force acting on the box is
According to Newton's second law of motion, the net force on an object is equal to the product between its mass (m) and its acceleration (a), so we can write:
And so
However, in this case the box is moving at constant speed; this means that its acceleration is zero:
Therefore we have:
Which means
And since we are told that
This means that the force of friction is also 25 N: