Question

Three forces act simultaneously on a 2.70-kg block. For clarity the drawings show the forces separately. The block starts from rest and, as a result of the combined action of the forces, undergoes a displacement s, which has a magnitude of 6.50 m. What is the final speed of the block

Answer 1

The image depicting the 3 forces is missing, so I've attached the image.

Answer:

v = 27.09 m/s

Explanation:

If we assume that no friction exists, we can solve as follows;

From the images we see that;

First image: F=75N acting at angle of 38°

Second Image: F = 54N acting Horizontally

Third Image: F=93N acting at 65°

Thus,

Sum of forces about the x - axis

ΣFx = 75cos(38) + 54cos(0) + 93cos(65)

ΣFx = (75 x 0.7880) + (54 x 1) + (93 x 0.4226)

ΣFx = 152.4 N

Now,

Sum of the forces =mass x acceleration,

Thus,

ΣFx = ma

Mass is given as 2.7kg

Thus,

152.4 = 2.7a

a = 152.4/2.7

a = 56.44 m/s²

Using Newton's equation of motion;

v² = u² + 2as

Where;

v is final velocity

u is initial velocity

a is acceleration

s is distance traveled in the x direction

Now, it starts from rest, thus, u = 0m/s

a =56.44 m/s²

s = 6.5m

Thus;

v² = 0 + 2(56.44)(6.5)

v² = 733.72

v = √733.72

v = 27.09 m/s