## anonymous 4 years ago The magnitude of each force is 286 N, the force on the right is applied at an angle 29 and the mass of the block is 20 kg. The coefficient of friction is 0.152. The acceleration of gravity is 9.8 m/s2 .

1. anonymous

What is the magnitude of the resulting ac- celeration? Answer in units of m/s2

2. TuringTest

needs a picture...

3. anonymous

ok

4. anonymous

|dw:1328461258569:dw|

5. anonymous

|dw:1328461435693:dw|

6. anonymous

then n * coefficient of friction which in this case is .152 would be f(k) = .874

7. anonymous

|dw:1328461609676:dw|

8. anonymous

i need to know if this is right!!!! please.

9. anonymous

no one??????

10. TuringTest

I get 26.6m/s^2

11. TuringTest

$\sum F_y=F\sin\theta-F_g=286\sin(27^{\circ})-20(9.8)\approx-57.3$$N\approx57.3$$\sum F_x=F+F\cos\theta-f\to a=\frac1m[F(1+\cos\theta)-\mu N]$$=\frac1{20}[286(1+\cos(27^{\circ}))-0.152(57.3)] \approx26.6m/s^2$

12. anonymous

where did the 27 degrees come from???

13. TuringTest

typo, meant 29 I can't see your answer so I couldn't check it

14. TuringTest

...I actually switched numbers partway through my calculation$\sum F_y=F\sin\theta-F_g=286\sin(29^{\circ})-20(9.8)\approx-57.3$$N\approx57.3$$\sum F_x=F+F\cos\theta-f\to a=\frac1m[F(1+\cos\theta)-\mu N]$$=\frac1{20}[286(1+\cos(29^{\circ}))-0.152(57.3)] \approx26.37m/s^2$