Question 11
First sentence is meant for calculation of k
k=5000 Nm*-1
When the object come to a rest , it has fallen (10+x) cm where x is the compression of the spring.
Initial KE and Final KE = 0, hence EPE1 + PE1 = EPE2 + PE2
0 + mg(0.10 +x)= 0.5k x*2
x = 0.086 m
Height of spring = 0.20 - 0.086 = 0.114 m
Question 12
KE1 + PE1 = KE2 + PE2 + W
KE1 + 0 = 0 + 0 + Fd
F = KE/d = 2500 N
Question 13
a) W = Fd = 1.00 x 10*12 J
b) KE1 + PE1+ W = KE2 + PE2
KE1 + 0 + 1.00*12 = KE2 + 0
KE2 = 4.025 x 10*12
v = 1.27 x 10*4 ms*-1
Question 15
KE1 + PE1 = KE2 + PE2 + W
Note : W = Fd where d= 57 is the length of the slope and F the frictional force
PE2 = 0 and PE 1 = mgh where h =57sin 25
Ans: v = 18.7 ms*-1
