*Only two horizontal forces act on a 3.0 kg body that can move over a frictionless floor. One force is 9.0 N, acting due east, and the other is 8.0 N, acting 62° north of west. What is the magnitude of the body's acceleration?*

Answer in the book (with no workings) says 2.9m/s^{2}

I get ~1.7m/s^{2}

F(net,x) = ma

F1(x) = 9.0N

F2(x) = -8.0cos(62) = ~ -3.8N

F(net,x) = F1(x) + F2(x) = 9.0N - 3.8N = 5.2N

5.2N = 3.0kg * a(m/s^{2})

a = 5.2N / 3.0kg = ~1.7m/s^{2}

Seems like a straight forward question, what am I doing wrong?

Guest Mar 12, 2017

#1**+5 **

the forces acting laterally are

+9 N to the east

and -8cos62 = -3.755 N

Net force is +5.244 east f= ma 5.244 = 3 kg a a = 1.75 m/s^2 east

Either the book has the wrong answer or the wrong info in the question!

Good job !

ElectricPavlov
Mar 12, 2017

#2**0 **

..as an 'aside' if the angle is 6.2 degrees north of west, the answer comes out to a = 2.87 m/s^2

Maybe there is a typo in the angle of the 8N force ??

ElectricPavlov
Mar 12, 2017