Question

A cord is used to vertically lower an initially stationary block of mass M = 3.6 kg at a constant downward acceleration of g/7. When the block has fallen a distance d = 4.2 m, find (a) the work done by the cord's force on the block, (b) the work done by the gravitational force on the block, (c) the kinetic energy of the block, and (d) the speed of the block. (Note : Take the downward direction positive)

Answer 1

(a)
`W_c=127.008 J`

(b)
`W_g=148.176 J`

(c) K.E. = 21.168 J

(d)
`v=3.4293m.s^(-1)`

Step-by-step explanation:

Given:

• mass of a block, M = 3.6 kg

• initial velocity of the block,
  `u=0 m.s^(-1)`

• constant downward acceleration,
  `a_d= (g)/(7)`

`\Rightarrow` That a constant upward acceleration of
`(6g)/(7)` is applied in the presence of gravity.

∴
`a=- (6g)/(7)`

• height through which the block falls, d = 4.2 m

(a)

Force by the cord on the block,

`F_c= M* a`

`F_c=3.6* (-6)*(9.8)/(7)`

`F_c=-30.24 N`

∴Work by the cord on the block,

`W_c= F_c* d`

`W_c= -30.24* 4.2`

We take -ve sign because the direction of force and the displacement are opposite to each other.

`W_c=-127.008 J`

(b)

Force on the block due to gravity:

`F_g= M.g`

∵the gravity is naturally a constant and we cannot change it

`F_g=3.6* 9.8`

`F_g=35.28 N`

∴Work by the gravity on the block,

`W_g=F_g* d`

`W_g=35.28* 4.2`

`W_g=148.176 J`

(c)

Kinetic energy of the block will be equal to the net work done i.e. sum of the two works.

mathematically:

`K.E.= W_g+W_c`

`K.E.=148.176-127.008`

K.E. = 21.168 J

(d)

From the equation of motion:

`v^2=u^2+2a_d* d`

putting the respective values:

`v=\sqrt{0^2+2* (9.8)/(7)* 4.2 }`

`v=3.4293m.s^(-1)` is the speed when the block has fallen 4.2 meters.