The coefficients of static and kinetic friction between a 50.0-kg box and a horizontal surface are 0.500 and 0.400 respectively. (a) What is the acceleration of the object if a 250-N horizontal force is applied to the box

1.08 m/s^2

Explanation:

The maximum value of static friction F_s = μ_s mg

where μ_s = coefficient of static friction

m= mass of box= 50kg

g= 9.81 m/s^2 , acceleration due to gravity

F_s = 0.5×50×9.81 =245 N

Since, the applied force is greater than the maximum value of static friction F_s, the object must be in motion. Hence, kinetic friction must be taken into account and not static friction.

By equilibrium condition

ma = F - μ_k m g

where F = 250 N,  μ_k = coefficient of kinetic friction = 0.4, m= 50 Kg

a = acceleration of the box

50a = 250-0.4×50×9.81

Solving we get a=  1.08 m/s^2

wind moving at higher speeds will transfer more energy to the water, resulting in stronger waves.

0.19 m/s and 0.45 m/s