How do I calculate work in two-dimensions and find the final velocity and time?

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In summary, the conversation discusses a physics problem involving calculating force, work, velocity, acceleration, and time. The problem statement provided conflicting information about the distance, but the individual asking the question solved it using the given 15m. The problem may be hinting at the law of cosines with respect to the 35° slope. The answer provided by the individual was correct, but had too many significant figures. They also acknowledged that the question may be a recycled one with incorrect data.
  • #1
Edel Crine
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Homework Statement
Starting from rest, an intern pushes a 45-kg gurney 40 m down the hall with a constant force of 80 N directed downward at an angle of 35° with respect to the horizontal.
(a) What is the work done by the intern on the gurney during the 15-m trip?
(b) How fast is the gurney going when it has moved 15 m?
(c) How much time elapses during the 15-m journey? Ignore friction.
Relevant Equations
W=F*x
I'm not sure that does the given distance represent "x" direction? If so, I did...

a) Fx = 80N * cos(35°) = 65.53N
W = F*Δx = 65.53N * (15m) = 982.98 J

b) k = 982.98 J = 1/2 (m)(v2)
v = (√2k/m) = 6.6 m/s

c) vf = vi + at

a = F/m = 1.46 m/s2

t = (vf-vi)/a = 4.54 sec

But I'm not sure for I got... h
 
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  • #2
The problem statement says 40m, but then asks about 15m. Which is it?
 
  • #3
haruspex said:
The problem statement says 40m, but then asks about 15m. Which is it?
Well, the problem statement said 40m, but subproblems said 15m, so I think we need to solve it by 15m?
 
  • #4
I think that the problem may be hinting at the law of cosines regarding the 35° slope.
 
  • #5
sysprog said:
I think that the problem may be hinting at the law of cosines with respect to the 35° slope.
yeah, I think so too and that's how I did...?
 
  • #6
Edel Crine said:
Well, the problem statement said 40m, but subproblems said 15m, so I think we need to solve it by 15m?
Sounds to me like a recycled problem. The question setter failed to update all the parts.
You have answered correctly for 15m, except that you quote too many significant figures. The given data only provides two.
 
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  • #7
Edel Crine said:
yeah, I think so too and that's how I did...?
It looks that way to me, but I think that @haruspex and others here are more expert on this than I am .
 
  • #8
haruspex said:
Sounds to me like a recycled problem. The question setter failed to update all the parts.
You have answered correctly for 15m, except that you quote too many significant figures. The given data only provides two.
Oh yes, I almost forgot about it...hh 'Appreciate!
 
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  • #9
sysprog said:
It looks that way to me, but I think that @haruspex and others here are more expert on this than I am .
You are good! I appreciate every comment from you hh
 
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1. How do I calculate work in two-dimensions?

To calculate work in two dimensions, you will need to use the formula W = Fdcosθ, where W is work, F is the force applied, d is the displacement, and θ is the angle between the force and displacement vectors.

2. How do I find the final velocity in two-dimensions?

To find the final velocity in two dimensions, you will first need to calculate the net force acting on the object using the formula F = ma, where m is the mass of the object and a is the acceleration. Then, use the formula vf = vi + at, where vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time.

3. How do I find the time in two-dimensions?

To find the time in two dimensions, you can use the formula t = (vf - vi)/a, where t is the time, vf is the final velocity, vi is the initial velocity, and a is the acceleration. Alternatively, if you know the displacement and average velocity, you can use the formula t = d/vavg.

4. What units should I use when calculating work and velocity in two-dimensions?

When calculating work, the units should be in joules (J), which is equal to newtons (N) multiplied by meters (m). When calculating velocity, the units should be in meters per second (m/s).

5. Can I use these equations for any type of motion in two-dimensions?

Yes, these equations can be used for any type of motion in two-dimensions, as long as the motion is constant and there are no external forces acting on the object. If the motion is not constant or there are external forces, more complex equations may need to be used.

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