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

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

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

the independent variable, the one that is intentionally manipulated, is whether or not the car hits the barrier.

the dependent variable, the one that you measure the response in, is the distance that the washers travel.