A 7.8 kg block initially at rest is pulled to the
right along a horizontal, frictionless surface
by a constant, horizontal force of 14.8 N.
Find the speed of the block after it has
moved 2.8 m.
Answer in units of m/s
Please identify the School Subject carefully. 11th grade will not get the right teacher reading your post.
Sra
To find the speed of the block after it has moved 2.8 m, we can use the concept of work and energy. The work done on an object is equal to the change in its kinetic energy.
First, we need to find the work done on the block. Work is given by the formula:
Work = Force x Distance x cos(θ)
In this case, the force applied is 14.8 N, and the distance is 2.8 m. Since the force is applied horizontally, the angle θ between the force and the direction of motion is 0 degrees. Therefore, cos(0) = 1.
Work = 14.8 N x 2.8 m x 1 = 41.44 J
Now, we can calculate the change in kinetic energy. The work done on the block is equal to the change in kinetic energy:
Work = Change in Kinetic Energy
41.44 J = (1/2)mv^2 - (1/2)m(0)^2
Since the block is initially at rest, the initial kinetic energy is zero. Therefore, we can simplify the equation:
41.44 J = (1/2)mv^2
Now, we can solve for the speed (v):
v^2 = (41.44 J) x 2 / (1 kg x 7.8 kg)
v^2 = 10.56 m^2/s^2
Taking the square root of both sides:
v = √(10.56 m^2/s^2)
v = 3.25 m/s
Therefore, the speed of the block after it has moved 2.8 m is 3.25 m/s.