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    Hi, I am a bit stuck on this mechanics questions:
    A truck accelerates from rest down a slope that makes an angle of 30 degrees to the horizontal. What is the distance traveled by the truck after 4 seconds?

    I used the suvat equation, taking acceleration due to gravity as 9.81ms-2

    s= ut + 0.5at^2
    I got 78.48 as an answer. This is one of the options of multiple choice. However, I was wondering isn't this distance the vertical distance? Do you not have to calculate the distance down the slope by doing

    78.84/sin(30)
    However, this answer isnt in the multiple choice answer so why do you not have to do this step? Why is the distance only the vertical distance
    Thank you to anyone who helps! Will rep best answer
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    You cannot take acceleration to be simply the strength of gravity, because this ignores the reaction force imparted on the truck by the slope.

    Try splitting gravity into two component forces which are respectively parallel and perpendicular to the slope. The perpendicular component will be cancelled out by the reaction force, leaving only the parallel component which will accelerate the truck downhill along the slope.
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    (Original post by Lola1244)
    Hi, I am a bit stuck on this mechanics questions:
    A truck accelerates from rest down a slope that makes an angle of 30 degrees to the horizontal. What is the distance traveled by the truck after 4 seconds?

    I used the suvat equation, taking acceleration due to gravity as 9.81ms-2

    s= ut + 0.5at^2
    I got 78.48 as an answer. This is one of the options of multiple choice. However, I was wondering isn't this distance the vertical distance? Do you not have to calculate the distance down the slope by doing

    78.84/sin(30)
    However, this answer isnt in the multiple choice answer so why do you not have to do this step? Why is the distance only the vertical distance
    Thank you to anyone who helps! Will rep best answer
    Hey. I know what question you're on about (have the book too). The answer is D, 19.6m. How i got to this answer is as followed....
    First i considered that the resultant force (F of F=ma) is equal the component of the truck's weight that is parallel to the slope inclined at 30 degrees. Therefore i got the the expression Sin(30)xW. Since Sin(30)xW is equal to F=ma, i expressed this as an equation--> Sin(30)xW=MA. W=mg, therefore the original equation made can be expressed as (which i done)--> Sin(30)xMG=MA. G=9.81--> Sin(30)xM(9.81)=MA. The next step i took was to divide by M in order to make A (acceleration) the subject, giving--> Sin(30)x(9.81)=A. Therefore A=4.905. With acceleration calculated and the time being given, i was able to calculate the distance traveled--> 4.905x4=19.62m-->19.6m.

    PS. If my explanation wasn't clear enough, ask through personal message and i'll answer ASAP .
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    Just to let you know this is a duplicate thread

    http://www.thestudentroom.co.uk/show....php?t=3694191
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    (Original post by newfanta123)
    Hey. I know what question you're on about (have the book too). The answer is D, 19.6m. How i got to this answer is as followed....
    First i considered that the resultant force (F of F=ma) is equal the component of the truck's weight that is parallel to the slope inclined at 30 degrees. Therefore i got the the expression Sin(30)xW. Since Sin(30)xW is equal to F=ma, i expressed this as an equation--> Sin(30)xW=MA. W=mg, therefore the original equation made can be expressed as (which i done)--> Sin(30)xMG=MA. G=9.81--> Sin(30)xM(9.81)=MA. The next step i took was to divide by M in order to make A (acceleration) the subject, giving--> Sin(30)x(9.81)=A. Therefore A=4.905. With acceleration calculated and the time being given, i was able to calculate the distance traveled--> 4.905x4=19.62m-->19.6m.

    PS. If my explanation wasn't clear enough, ask through personal message and i'll answer ASAP .
    Hiya! Thank you so much for your reply! I understand how you have split up the acceleration but isnt s= ut + 1/2 =at^2 ... so you need to do 4.905 x 16 surely? Which gives you 39.2
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    (Original post by Lola1244)
    Hiya! Thank you so much for your reply! I understand how you have split up the acceleration but isnt s= ut + 1/2 =at^2 ... so you need to do 4.905 x 16 surely? Which gives you 39.2
    Oops...i realize now that i've just worked out the acceleration (Sin(30)x9.81) and final velocity (19.6). So you're right. Sorry about that
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    (Original post by newfanta123)
    Oops...i realize now that i've just worked out the acceleration (Sin(30)x9.81) and final velocity (19.6). So you're right. Sorry about that
    Hahah don't worry! Have you done the question after this one- no5. where two boxes A and B are connected by a light thread called X. A is 2kg whereas B is 3kg and a frictional force of 10 N acts on each box. A force of 10N is applied to string Y which is connected to Box B. The boxes are moving at constant velocity of 2 m/s. And it asks which of the following statements are true:

    F is slightly greater than 20N- this is false because F must be equal to 20N as this is what Newton's First Law tells us
    String X exerts a force of 10N on B- now here is where I get confused. String X must be exerting a force on B which is in the opposite direction to the motion of the travel of the boxes. Does the frictional force of A act along this string? This would mean that X did exert a force of 10N on B. However, I thought friciton acted between the boxes and the ground?
    X exerts a greater force on A than it does on B
    X exerts a greater force on B than it does on A

    I am so confused I would be very grateful if you could help out
 
 
 
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