Apendulum has a mass of 1.5 kg and starts at a height of 0.4 m. if it is released from rest, how fast is it going when it reaches the lowest point of its path? acceleration due to gravity is g = 9.8 m/s2. a. 2.8 m/s b. 0 m/s c. 5.9 m/s d. 4.3 m/s

A. 2.8 m/s

Explanation:

Suppose that at the height of 0 m, the path of the pendulum is lowest.

If we use law of conservation of energy, the pendulum will have zero kinetic energy or K.E when it is at highest point, because K.E happens during movement of object and at the highest point all the energy will be P.E

                                                    P.E= mgh

Similarly, when the pendulum reaches at the lowest point, the height becomes zero and the P.E also becomes zero. Now all the energy will be K.E

                                               K.E= 1/2 m v^2

In question, we are asked about the speed as the pendulum  it reaches the lowest point of its path. Like we mentioned P.E will be zero at lowest point  because of zero height. And also we will use law of conservation of energy because no energy has been lost from system.

                                                  K.E=     P.E

                                       1/2 m v^2  =   mgh

Taking sq.root at both sides

                                                  v= Under root 2 gh

                                                   v=Under root 2x 9.8 m/s x0.4 m

                                                   v=Under root 7.84

                                                    v=2.8 m/sec

Hope it helps!

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg