How long does it take for the toboggan to reach the bottom of the slope?

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SUMMARY

The discussion centers on calculating the time it takes for a toboggan, initially carrying 35 kg of sand, to slide down a 90m slope inclined at 30 degrees. The toboggan's mass decreases due to sand leaking at a rate of 2.3 kg/s. Initially, the user calculated the time to be approximately 6.061 seconds using a constant acceleration of 0.5g. After considering the effect of the sand exiting the toboggan, the revised time was calculated to be about 5.105 seconds.

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Sekonda
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Hey guys, here is the question :

A 5.4kg toboggan carrying 35kg of sand slides from rest down an icy 90m slope inclined
at 30◦ below the horizontal. The sand leaks from the back of the toboggan at a rate of
2.3kgs−1. How long does it take the toboggan to reach the bottom of the slope?

I found that the mass of the system at anyone time can be written as a simple linear relationship and so the weight of the system down the slope at any time 't' can also be expressed using this same relationship.

Using this to determine the acceleration down the slope at any time 't', I found that the acceleration is constant at 0.5g down the slope.

Using the SUVAT equations I then determined the time taken to be approximately 6.061s.

Is this correct? Or am I being an idiot somewhere :P

Thanks for any comments/help
Cheers!
 
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The fact that the sand is exiting parallel to the direction of motion will add acceleration to the sled.
 
Last edited:
Ahh yes, I thought I had oversimplified the problem. I factored that in and got the time as about 5.105s.

Cheers man!
 

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