a block of mass 2kg moves on a frictionless inline (none of the forces on the block are shown). its initial velocity is v initial=3m/s, and it slides a vertical distance h=10m up the slope and its final velocity is v final=2m/s. find the external work done on the block.
so is it 196 or 191J?
To find the external work done on the block, we can use the work-energy principle. According to the principle, the work done on an object is equal to the change in its kinetic energy.
The work done by external forces can be calculated by the equation:
Work done (W) = Change in kinetic energy (ΔKE)
The change in kinetic energy can be determined using the formula:
ΔKE = KE final - KE initial
Let's break down the problem to find the necessary values:
Given:
Mass of the block (m) = 2 kg
Initial velocity (v initial) = 3 m/s
Final velocity (v final) = 2 m/s
Vertical distance (h) = 10 m
First, we calculate the initial kinetic energy:
KE initial = (1/2) * m * (v initial)^2
Substituting the given values:
KE initial = (1/2) * 2 kg * (3 m/s)^2 = 9 J
Next, we calculate the final kinetic energy using the final velocity:
KE final = (1/2) * m * (v final)^2
Substituting the given values:
KE final = (1/2) * 2 kg * (2 m/s)^2 = 4 J
Now we can find the change in kinetic energy:
ΔKE = KE final - KE initial
ΔKE = 4 J - 9 J
ΔKE = -5 J
Since the change in kinetic energy is negative, it indicates that work is done against the block. Therefore, the external work done on the block is 5 Joules (magnitude only).
Note: Since the problem mentions that there are no frictional forces, the only external force is gravity acting vertically downwards.
initial PE+InitialKE+Workdone=finalPE+finalKE
0+1/2 m 3^2 + workdone=mg*10 + 1/2 m 2^2
solve for work done.