A 2-kg block is dragged over a rough horizontal surface by a constant force of 15 N acting at an angle of 60° above the horizontal. The speed of the block increases from 4.0 m/s to 8 m/s in a displacement of 9 m. What work was done by the friction force during this displacement?
Need quick help please
Net horizontal force = mass *acceleration = 2 a
goes 9 meters in time t
v = Vi + a t = 4 + at = 8
so
a t = 4 and a = 4/t
distance= 4 t + (1/2) a t^2 = 9
4 t + (1/2)(4/t) t^2 = 9
4 t + 2 t = 9
t = 1.5 seconds
a = 4/1.5 = 2.67 m/s^2
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net horizontal force = m a = 2 * 2.67 = 5.34 Newtons
work done by net horizontal force = 5.34 * 9 = 48.06 Joules
work done by the puller = 15 cos 60 * 9 = 67.5 Joules
so
difference is work turned to heat by friction 67.5 - 48.1
A 2-kg block is dragged over a rough horizontal surface by a constant force of 12 N acting at an angle of 60° above the horizontal. The speed of the block increases from 4.0 m/s to 11 m/s in a displacement of 9 m. What work was done by the friction force during this displacement?
To find the work done by the friction force, we need to calculate the net work done on the block.
The net work done is equal to the change in kinetic energy of the block. The change in kinetic energy is given by the equation:
ΔKE = KE_final - KE_initial
First, let's find the initial kinetic energy (KE_initial) of the block. The equation for kinetic energy is:
KE = (1/2) * m * v^2
where m is the mass of the block and v is its velocity.
We're given that the mass of the block (m) is 2 kg and the initial velocity (v_initial) is 4.0 m/s. Plugging these values into the equation, we have:
KE_initial = (1/2) * 2 kg * (4.0 m/s)^2
KE_initial = 16 J
Next, let's find the final kinetic energy (KE_final) of the block. Given that the final velocity (v_final) is 8 m/s, we can plug this value into the equation for kinetic energy:
KE_final = (1/2) * 2 kg * (8 m/s)^2
KE_final = 64 J
Now, we can find the change in kinetic energy:
ΔKE = KE_final - KE_initial
ΔKE = 64 J - 16 J
ΔKE = 48 J
The change in kinetic energy (ΔKE) is equal to the work done on the block. Therefore, the work done by the friction force during this displacement is 48 Joules.