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a block of mass m 5.00 kg is

pulled along a horizontal frictionless floor by a cord

that exerts a force of magnitude F 12.0 N at an angle

u 25.0°. (a) What is the magnitude of the

block’s acceleration? (b) The force magnitude F is

slowly increased. What is its value just before the

block is lifted (completely) off the floor? (c) What is

the magnitude of the block’s acceleration just before

it is lifted (completely) off the floor?

Verified Answer :

a.
the force in the direction of movement
Fx = F x cosA

Fx = 12 x cos25

Fx = 10.88

m x a = 10.88

a = 10.88/5
a= 2.18 m/s^2

b.
when the block will be lifted off the floor and when the vertical component
Fy = mg

Fy = F x sin25*
Fy= 5 x 9.8

Fy= 115.94 N

c.
if Fx = ma

a = 115.94/5
a= 23.19 m/s^2
hope it helps

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