Work = force x distance
200 Newtons x 20 meters
= 4,000 Joules
Final velocity = 0, thus final kinetic energy is 0
Initial kinetic energy:
0.5mv²
= 0.5 x 0.48 x 4.08²
= 4.0 J
Decrease in kinetic energy = 4 - 0 = 4 Joules
Answer:
24 Coulumbs
Explanation:
Given data
time= 1 minute= 6 seconds
P=2 W
R= 12 ohm
We know that
P= I^2R
P/R= I^2
2/12= I^2
I^2= 0.166
I= √0.166
I= 0.4 amps
We know also that
Q= It
substitute
Q= 0.4*60
Q= 24 Columbs
Hence the charge is 24 Coulumbs
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Answer:
A) 89.39 J
B) 30.39J
C) 23.8 J
Explanation:
We are given;
F = 30.2N
m = 3.5 kg
μ_k = 0.646
d = 2.96m
ΔEth (Block) = 35.2J
A) Work done by the applied force on the block-floor system is given as;
W = F•d
Thus, W = 30.2 x 2.96 = 89.39 J
B) Total thermal energy dissipated by the whole system which includes the floor and the block is given as;
ΔEth = μ_k•mgd
Thus, ΔEth = 0.646 x 3.5 x 9.8 x 2.96 = 65.59J
Now, we are given the thermal energy of the block which is ΔEth (Block) = 35.2J.
Thus,
ΔEth = ΔEth (Block) + ΔEth (floor)
Thus,
ΔEth (floor) = ΔEth - ΔEth (Block)
ΔEth (floor) = 65.59J - 35.2J = 30.39J
C) The total work done is considered as the sum of the thermal energy dissipated as heat and the kinetic energy of the block. Thus;
W = K + ΔEth
Therefore;
K = W - ΔEth
K = 89.39 - 65.59 = 23.8J
