A dockworker applies a constant horizontal force of 82.0 to a block of ice on a smooth horizontal floor. The frictional force is negligible. The block starts from rest and moves a distance 12.5 in a time 4.60 .
I suppose you want the mass?
82 = m a
d = (1/2)a t^2
12.5 = .5 a (4.6)^2
a = 1.18
so
82 = m (1.18)
m = 69.4 kg
I have to assume that you are using SI units. Please in the future type the entire problem including units and the question itself.
To find the acceleration of the block, which is required to calculate the force applied by the dockworker, we can use the equation of motion:
s = ut + (1/2)at^2
Where:
s = distance = 12.5 m
u = initial velocity = 0 m/s (as the block starts from rest)
t = time = 4.60 s
Rearranging the equation, we get:
a = 2s/(t^2)
Plugging in the values, we get:
a = 2 * 12.5 m / (4.60 s)^2
Calculating this, we find:
a ≈ 0.791 m/s^2
Therefore, the acceleration of the block is approximately 0.791 m/s^2.
Now, to calculate the force applied by the dockworker, we can use Newton's second law of motion:
F = ma
Where:
F = force applied (to be determined)
m = mass of the block (unknown)
a = acceleration = 0.791 m/s^2
Since the block is on a smooth horizontal floor, the only horizontal force acting on it is the force applied by the dockworker. Therefore, this force is equal to the mass of the block times its acceleration:
F = ma
We don't have the mass of the block given in the problem, so we cannot calculate the force applied by the dockworker without knowing the mass.