A 8.35 kg block initially at rest is pulled to the right along a horizontal, fri

Question

A 8.35 kg block initially at rest is pulled to the right along a horizontal, frictionless surface by a constant horizontal force of 6.4 N. Find the block's speed after it has moved7.7 m.

The result for first square is 3.44 but for the Master it section I have no idea...

A Block Pulled on a Frictionless Surface A 8.35 kg block initially at rest is pulled to the right along a horizontal, frictionless surface by a constant horizontal force of 6.4 N. Find the block's speed after it has moved 7.7 m. SOLVE IT Conceptualize: The figure illustrates this situation. Imagine pulling a toy car across a table with a horizontal rubber band attached to the front of the car. The force is maintained constant by ensuring that the stretched rubber band always has the same length A block on a frictionless surface is pulled to the right by a constant horizontal force Categorize: We could apply the equations of kinematics to determine the answer, but let us practice the energy approach. The block is the system, and three external forces act on the system. The normal force balances the gravitational force on the block, and neither of these vertically acting forces does work on the block because their points of application are horizontally displaced Analyze: The net external force acting on the block is the horizontal 6.4 N force Find the work done by this force on the block: W = Fax = (6.4 N)(7.7 m) = 49.28 J Use the work-kinetic energy theorem for the block and note that its initial kinetio energy is zero: Solve for vyi -2W m/s 8.35 kg Finalize: It would be useful for you to solve this problem again by modeling the block as a particle under a net force to find its acceleration and then as a particle under constant acceleration to find its final velocity MASTER IT HINTS: GETTING STARTED I IMSTUCK! For the above example, suppose that the magnitude of the force is doubled. Through what displacement will the block travel before reaching the same final speed found above?

Explanation / Answer

In both cases Workdone is same. Because the change in kinetic energy is same.

so, the product of F*delta_x is same

so, When F becomes double delta_x becomes half.

so, delta_x = 7.7/2

= 3.85 m <<<<<<---------------Answer

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