Question 18
a) KE1 + PE1+ W = KE2 + PE2
0 + 0 + W = 7.5 x 10*4 + 0
W = 7.5 x 10*4 J
b) Power = W/t = 2.5 x 10*4 W
c) P(at t=3) = F v(at t=3)= ma(10) = 5 x 10*4W
Note : Average power = total energy divide by time taken
Instantaneous power = force x velocity (at that instant)
Question 19
KE = PE = mgh =1.177x 10*9 J
Electrical energy derived from KE in 1 sec = 0.75 (1.177 x 10*9)= 8.83x10*8J
Electrical Power = 8.83x1089 W (energy in 1 sec)
Question 20
P = Fv where F = kv*2
P= kv*3
Question 21
Energy per day= 8 kWh
Energy per month= 248 kWh
Cost = $62
Question 22
P = Fv but F = kv
Hence P = kv*2
P1/P2 = v1*2/v2*2
P2= 7500x (12/4)*2
= 6.75 x10*4W
