# Forces Help

#### omonoid

Part 1

1. The only two forces acting on a body have magnitudes of 20 N and 42 N and directions that differ by 60°. The resulting acceleration has a magnitude of 20 m/s2. What is the mass of the body?

2. F=ma

3. Im not sure how to solve this because i'm used to being able to use trig like cos and tan, but this problem doesn't create a triangle with 90 degrees

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Part 2

1. A bullet of mass 6.5*10-3 kg moving at 1150 m/s impacts with a large fixed block of wood and travels 8 cm before coming to rest. Assuming that the deceleration of the bullet is constant, find the force exerted by the wood on the bullet.

2. F=ma
V^2=Vi^2+2ax

3. V=0 Vi=1150 m/s x=.08 m solve for a and i got -8265625 m/s^2 then multiply by mass (6.5E-3) and get -53726.5625. Is this right? would i void the negative on a force?

Last edited:

#### alphysicist

Homework Helper
Hi omonoid,

Part 1

1. The only two forces acting on a body have magnitudes of 20 N and 42 N and directions that differ by 60°. The resulting acceleration has a magnitude of 20 m/s2. What is the mass of the body?

2. F=ma
Let me be picky here and say you should write this as Fnet=ma, that is, the sum of the forces equals m*a. Does that help? Have you added vectors together that were not perpendicular to each other?

3. Im not sure how to solve this because i'm used to being able to use trig like cos and tan, but this problem doesn't create a triangle with 90 degrees

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Part 2

1. A bullet of mass 6.5*10-3 kg moving at 1150 m/s impacts with a large fixed block of wood and travels 8 cm before coming to rest. Assuming that the deceleration of the bullet is constant, find the force exerted by the wood on the bullet.

2. F=ma
V^2=Vi^2+2ax

3. V=0 Vi=1150 m/s x=.08 m solve for a and i got -8265625 m/s^2 then multiply by mass (6.5E-3) and get -53726.5625. Is this right? would i void the negative on a force?
That number looks right to me. The minus sign is showing that the force is in the opposite direction of the initial velocity.

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