Aforce f = 94.9 n acts at an angle α = 35◦ with respect to the horizontal on a block of mass m = 29.3 kg, which is at rest on a horizontal plane. the acceleration of gravity is 9.81 m/s 2 . if the static frictional coefficient is µs = 0.69, what is the force of static friction? 29.3 kg µs = 0.69 94.9 n 35◦ 1. 144.69 n 2. 160.77 n 3. 37.5584 n 4. 53.6389 n 5. 198.329 n 6. 0 n 7. 235.887 n 8. 77.7375 n 9. 251.968 n

8) 77.7375 N

Explanation:

Given that

F = 94.9 N

α = 35◦

m = 29.3 kg

µs = 0.69

The component of force F

F1= 94.9 sin35◦ =54.43 N

F2= 94.9 cos35◦ =77.73 N

So

N + F1= mg

N= 29.3 x 9.81 - 54.43

N=233 N

So the maximum static friction force

fr = µs .N

fr = 0.69 x 233

fr = 160.77 N

The applied horizontal force = F2

F2=77.73 N

The applied horizontal force is less than the maximum value of static friction.It means that the block will not move.The block will remain in the static position.

So the static friction fr= f2

Static friction = 77.73 N

8. 77.7375

i think the answer would be b

answer would be letter choice (a),   a ball resting on a shelf. would be another example of potential energy   >   stored energy > not moving.

a ball spinning on a finger   >   kinetic energy > moving energy

a ball rolling on the ground   >   kinetic energy   > moving energy

a   ball flying through the air   > kinetic energy   >   moving energy

hope that                                         : )