A dock worker applies a constant horizontal force of 79.0 N to a block of ice on a smooth horizontal floor. The frictional force is negligible. The block starts from rest and moves 11.0 m in the first 6.00 s.

Question

A dock worker applies a constant horizontal force of 79.0 N to a block of ice on a smooth horizontal floor. The frictional force is negligible. The block starts from rest and moves 11.0 m in the first 6.00 s.

What is the mass of the block of ice?

*I notice two very common complaints on this site, especially when dealing with Physics and math questions. 1) You showed no work! 2) What kind of math is your course based on?! My answers to both of these are simple. I need help because I DON'T KNOW WHAT I AM DOING....and...it is Algebra based Physics. PLEASE help me!

Use the fact that the work done equals the change in kinetic energy.

F X = (1/2) M V^2

X is the distance moved. F is the force. The final velocity (after 6 seconds)is V. Since it accelerates uniformly, the average velocity 11/6 m/s is half the final velocity.

V = 22/6 = 3.667 m/s

Now solve the first equation for M.

M = 2 F X/ V^2

Answer 1

To find the mass of the block of ice, we can use the work-energy principle, which states that the work done on an object is equal to the change in its kinetic energy. In this case, the work done by the dock worker is equal to the change in kinetic energy of the block.

The work done is given by the equation:

Work = Force x Distance (W = F X)

In this case, the force applied by the dock worker is 79.0 N and the distance moved by the block is 11.0 m. Therefore, the work done is:

Work = 79.0 N x 11.0 m = 869 N⋅m

The change in kinetic energy can be calculated using the equation:

Change in Kinetic Energy = (1/2) x Mass x Velocity^2 (ΔKE = (1/2)mv^2)

The final velocity of the block after 6.00 seconds is the average velocity, which is half of the distance traveled divided by the time interval:

Velocity = Distance/Time = 11.0 m / 6.00 s = 1.833 m/s

Let's substitute this value and the work done into the equation:

869 N⋅m = (1/2) x Mass x (1.833 m/s)^2

Now, solve for the mass (M):

Mass = (2 x Work) / (Velocity^2)

Mass = (2 x 869 N⋅m) / (1.833 m/s)^2

Mass = 2 (869 N⋅m) / (3.357 m^2/s^2)

Mass = 1160 kg

Therefore, the mass of the block of ice is approximately 1160 kg.

A dock worker applies a constant horizontal force of 79.0 N to a block of ice on a smooth horizontal floor. The frictional force is negligible. The block starts from rest and moves 11.0 m in the first 6.00 s.

189.5

To find the mass of the block of ice, we can use the work-energy principle, which states that the work done on an object is equal to the change in its kinetic energy. In this case, the work done by the dock worker is equal to the change in kinetic energy of the block.