A 34kg crate resting on a horizontal surface is pushed as shown. If the coefficient of static friction s between the surfaces is 0.25, what is the minimum force (Fs) required to start the motion? 1

a.
1332.80N

b.
83.30N

c.
8.50N

d.
136.00N
Heeelppp plzz​

D (not so sure, look in the explanation and judge yourself)

Explanation:

Assume that the acceleration due to gravity is 9.81 m/s^2 downwards

Minimum force required = 0.25 x 34 x 9.81

= 83.385 N

So technically the force need to be greater than 83.385 so the answer should be d, not really sure if it should be b though really depends on the acceleration due to gravity.

a 2,000 calorie based diet.

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft