• #1
TaylorHoward21
10
1
<< Mentor Note -- New poster has been reminded to use the Homework Help Template when posting schoolwork questions >>

A 25.0 kg sled is pulled (assume frictionless) with a force of 15.0 N at an angle of 20 degrees above the horizontal. (a.) How long will it take to pull the sled 25m? (b.) How fast will the sled be moving after it has been pulled 25.0m?


-V0x = V0sinθ
- X = X0 + V0xt
- V = V0 + at
- X = X0 + V0t + 0.5at2
- ∑Fx = max

I was able to solve for the x-acceleration, ax = 0.56m/s2 but I cannot figure out how to use the given forces and acceleration to solve for the time at 25m. Am I to assume V0 = 0?

This is a practice problem for a quiz that I cannot seem to crack. Please help.
 
Last edited by a moderator:

Answers and Replies

  • #2
14,293
8,337
I would assume that but of course you could solve for an arbitrary v0

So you have a triangle and the 25m would be on the hypotenuse.
 
  • #3
Delta2
Homework Helper
Insights Author
Gold Member
5,695
2,473
From the wording of the problem it is not clear (at least not to me)
if the sled is pulled along an incline that makes 20 degrees angle with the horizontal,

or

the sled is being pulled along the horizontal with a force that makes 20 degrees with the horizontal (and the vertical component of the force is balanced by the weight of the sled).

If I judge by your calculation of acceleration we have the second case (horizontal pull with a force that makes angle with the horizontal). Am I correct?
 
  • #4
14,293
8,337
Ah I think you are right Delta it’s a horizontal plane with the force at 20 degrees. The horizontal component provides the acceleration and the vertical component reduces the frictional force due to gravity. But since it’s frictionless we can ignore the vertical component.

If you allow for an arbitrary v0 then you can solve for t in the ##s=v0+1/2 a*t^2##
 
  • #5
TaylorHoward21
10
1
From the wording of the problem it is not clear (at least not to me)
if the sled is pulled along an incline that makes 20 degrees angle with the horizontal,

or

the sled is being pulled along the horizontal with a force that makes 20 degrees with the horizontal (and the vertical component of the force is balanced by the weight of the sled).

If I judge by your calculation of acceleration we have the second case (horizontal pull with a force that makes angle with the horizontal). Am I correct?

Yes, you are correct. Here is my diagram:
Ah I think you are right Delta it’s a horizontal plane with the force at 20 degrees. The horizontal component provides the acceleration and the vertical component reduces the frictional force due to gravity. But since it’s frictionless we can ignore the vertical component.

If you allow for an arbitrary v0 then you can solve for t in the ##s=v0+1/2 a*t^2##
So assuming V0 = 0 I find that time t = 9.4 seconds.
 
  • #6
14,293
8,337
I get 9.41 sec if v0 is zero.

##F_x = 15 N * cos(20) = 14.09 N##

##a = F_x / m = 14.09 / 25 = 0.564 m/s^2##

Using ##s = 1/2 * a*t^2##

and solving for ##t## I get: ##t = sqrt(2s/a) = sqrt(2*25 / 0.564) = 9.41 sec ##
 

Suggested for: Force and motion

Replies
4
Views
487
Replies
8
Views
330
Replies
16
Views
506
  • Last Post
Replies
4
Views
784
Replies
10
Views
440
  • Last Post
Replies
27
Views
358
  • Last Post
Replies
13
Views
457
Top