Acord is used to vertically lower an initially stationary block of mass m = 9.1 kg at a constant downward acceleration of g/9. when the block has fallen a distance d = 4.6 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)

(a) -364.64 J

(b) +410.2 J

(c) 45.58 J

(d) 3.17 m/s

Explanation:

We apply Newton's second law:

∑F = m*a (Formula 1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Data:

m= 9.1 kg

ay= g/9 m/s²

g = 9.81m/s² : acceleration due to gravity

W= m*g :  Block weight

W= m*g =9.1 kg*9.8 m/s²= 89.18 N

d= 4.6 m

Problem development

We apply the formula 1 to calculate the cord's force on the block,(T):

∑Fy = m*ay

W-T = m*g/9

W-T = W/9

T = W-W/9=(8/9) W= (8/9) *89.18 = 79.27 N

(a) Work done by the cord's force on the block (Wc)

Wc= - T*d =  - 79.27 N*4.6 m = - 364.64 J

The work ,Wc, is negative because the force (T) goes in the opposite direction to the block displacement.

(b)Work done by the gravitational force on the block(Wb)

Wb = W*d= 89.18 N*4.6 m = +410.2 J

The work ,Wb , is positive because the force (W) goes in the same direction as the block displacement

(c) kinetic energy of the block (K)

K =( 1/2)*m*v²  

v² = v₀²+2*a*d   ,v₀=0 , v²=2*a*d= 2*(g/9)* 4.6

K= ( 1/2)* 9.1* 2*(9.8/9)* 4.6 = 9.1* (9.8/9)* 4.6= 45.58 J

(d) the speed of the block.(v)

v²=2*a*d= 2*(g/9)* 4.6= 2*(9.8/9)* 4.6

v= 3.17 m/s

b.

explanation:

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