Question #141508

A small smooth object slides from rest down a smooth inclined plane inclined at 30 degrees to the horizontal. What is (i) the acceleration down the plane (ii) the time to reach the bottom if the plane is 5m long? The object is now thrown up the plane with an initial velocity of 15m/s. (iii) How long does the object take to come to rest? (iv) How far up the plane has the object then traveled?

Expert's answer


I) Find the acceleration down the plane:

a=g⋅sin(30∘)=9.8⋅12=4.9(m/s2).a=g\cdot sin(30^{\circ})=9.8\cdot\frac{1}{2}=4.9(m/s^2).

II) Find the time to reach the bottom if the plane is 5m long:

a⋅t22=L=5(m);\frac{a\cdot t^2}{2}=L=5(m);

t=L⋅2a=5⋅24.9=1.43(s).t=\sqrt{\frac{L\cdot2}{a}}=\sqrt{\frac{5\cdot2}{4.9}}=1.43(s).


III) Find the time the object take to come to rest:

v1=v0−a⋅t1=0;v_1=v_0-a\cdot t_1=0;

t1=v0a=154.9=3.06(s).t_1=\frac{v_0}{a}=\frac{15}{4.9}=3.06(s).

IV) Find how far up the plane has the object then traveled:

L=v0⋅t1−a⋅t122=L=v_0\cdot t_1-\frac{a\cdot t^2_{1}}{2}=

=15⋅3.06−4.9⋅3.0622=22.96(m).=15\cdot3.06-\frac{4.9\cdot 3.06^2}{2}=22.96(m).


Answer: I) 4.9(m/s2)4.9(m/s^2); II) 1.43(s)1.43(s); III) 3.06(s)3.06(s); IV) 22.96(m)22.96(m).


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