A 0.8 kg block of wood has an initial velocity of 0.25 m/s as it begins to slide across a table. The block comes to rest over a distance of 0.72m.
a) what is the average frictional force on the block?
b)How much work is done on the block by friction?
c)How much work is done on the table by the block?
But if i do this for force of friction, the result of frictional force is going to be negative. :(
avg acceleration=frictionforce/mass
Vf^2=Vi^2+2a*distance
Vf^2=Vi^2+2*frictionforce/mass * distance
solve for friction force.
work=force*distance
To find the average frictional force, we can use Newton's second law, which states that the force acting on an object is equal to its mass multiplied by its acceleration (F = m * a). In this case, the block of wood comes to rest, so its final velocity is 0 m/s.
a) To find the average frictional force on the block, we need to determine the acceleration first. We can use the equation of motion: v^2 = u^2 + 2as, where v is the final velocity (0 m/s), u is the initial velocity (0.25 m/s), a is the acceleration, and s is the distance covered (0.72 m). Rearranging the equation, we have a = (v^2 - u^2) / (2s).
Substituting the values, we get a = (0 - (0.25)^2) / (2 * 0.72) = -0.078 m/s^2 (the negative sign indicates the deceleration).
Now, we can calculate the average frictional force using the equation F = m * a. Given that the mass of the block is 0.8 kg, we get F = 0.8 kg * (-0.078 m/s^2) ≈ -0.0624 N.
Therefore, the average frictional force on the block is approximately -0.0624 N. The negative sign indicates that the frictional force opposes the direction of motion.
b) To determine the work done on the block by friction, we can use the equation W = F * d, where W is the work done, F is the force, and d is the distance over which the force is applied. In this case, the block comes to rest, so the work done by friction results in a loss of kinetic energy.
Since the block is initially moving and comes to rest, the work done on the block by friction is given by W = (1/2) * m * (v_f^2 - v_i^2), where m is the mass, v_f is the final velocity, and v_i is the initial velocity.
Here, m = 0.8 kg, v_f = 0 m/s, and v_i = 0.25 m/s. Substituting these values, we get W = (1/2) * 0.8 kg * (0 - (0.25)^2) = -0.03125 J.
Therefore, the work done on the block by friction is approximately -0.03125 J.
c) The work done on the table by the block is equal in magnitude but opposite in sign to the work done on the block by friction. Therefore, the work done on the table by the block is approximately 0.03125 J.