+0

# Force question

0
121
2

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/s2

I get ~1.7m/s2

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/s2)

a = 5.2N / 3.0kg = ~1.7m/s2

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

Guest Mar 12, 2017
Sort:

#1
+10657
+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
+10657
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

### 20 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details