Understanding the Dynamics of an Object on a Frictionless Surface

A 0.5 kg block is initially at rest on a frictionless horizontal surface. A force
F=10N is then applied horizontally to the block, as shown in the figure below. After the force is applied for 2 seconds, it is removed. Determine:

The acceleration of the block while the force is applied.
The velocity of the block after the force is removed.
The distance the block travels during the time the force is applied.
The total work done on the block by the force during the time it is applied.

    Title: Understanding the Dynamics of an Object on a Frictionless Surface Thesis Statement: By applying Newton's laws of motion and basic kinematic equations, we can determine the acceleration, velocity, distance traveled, and work done on a 0.5 kg block initially at rest on a frictionless horizontal surface when a constant force of 10 N is applied for 2 seconds. Acceleration of the Block: To calculate the acceleration of the block while the force is applied, we can use Newton's second law of motion, which states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. Therefore, the acceleration can be calculated as: [ a = \frac{F}{m} = \frac{10}{0.5} = 20 , m/s^2 ] Velocity of the Block after the Force is Removed: Using the kinematic equation ( v = u + at ), where ( v ) is the final velocity, ( u ) is the initial velocity (which is 0 m/s as the block was initially at rest), ( a ) is the acceleration calculated above, and ( t ) is the time for which the force is applied (2 seconds), we find: [ v = 0 + 20 \times 2 = 40 , m/s ] Distance Traveled by the Block: To calculate the distance traveled by the block during the time the force is applied, we can use the kinematic equation ( s = ut + \frac{1}{2}at^2 ), where ( s ) is the distance traveled. Since the initial velocity is 0 m/s, this simplifies to: [ s = \frac{1}{2}at^2 = \frac{1}{2} \times 20 \times (2)^2 = 40 , m ] Total Work Done on the Block by the Force: The work done on an object is given by the equation ( W = F \times s ), where ( W ) is work done, ( F ) is the force applied, and ( s ) is the distance moved in the direction of the force. Therefore, the total work done on the block during the time it is applied can be calculated as: [ W = 10 \times 40 = 400 , J ] In conclusion, by understanding fundamental principles of physics and utilizing relevant equations, we have successfully determined the acceleration, velocity, distance traveled, and total work done on a 0.5 kg block under the given conditions. These calculations showcase the interplay between forces, motion, and energy in a dynamic system.      

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