Determine the acceleration of the block

if the coefficient of static friction is 0.343.

a = 1.49m/s²

Explanation:

Find the diagram attached  

According to Newton's second law of motion

\sum Fx = ma

Fm - Ff = ma

Fm is the moving force

Ff is the frictional force

Fm = Wsin theta

Ff = μR = μWcos theta

Wsin theta - μWcos theta = ma

W is the weight = 50.1 * 9.8 = 490.98N

theta is the angle of inclination = 27.2°

μ is the coefficient of friction = 0.343

m is the mass = 50.1kg

a is the acceleration  

Substitute into the formula as shown;

Wsin theta - μWcos theta = ma

490.98sin27.2° - 0.343(490.98)cos 27.2° = 50.1a

224.423 - 149.78 = 50.1a

74.64 = 50.1a

a = 74.64/50.1

a = 1.49m/s²

Hence the acceleration of the block is a = 1.49m/s²

graphing motion: mastery test

209be

what is the instantaneous acceleration of

the particle at point b?

0 meters/second?

10 meters/second?

12 meters/second?

velocity imetersisecond

15 meters/second?

20

time (seconds)

answer: 0 meters/second^2