A small block slides along a path that is without friction until the block reaches the section L = 3m, which begins at height h=3m on a flat incline of angle 37°, as shown. In that section, the coefficient of kinetic friction is 0.50. The block passes through a point A with a speed of√136 m/s . Find the speed of the block as it passes through point B where the friction ends.
Answers
Answer:
4 m/s.
Explanation:
From the given question we get that the length is L = 3m and the height given is h=3m and the speed of the block when it passes the point A is vA = √136m/s with the angle of inclination of θ = 37°.
Again for the block we will have that mgsinθ + μ*mgcosθ = ma(where μ is the coefficient of friction)
Thus, the acceleration a = g(sin37 + cos 37)
Which on solving we will get a = 10 m/s^2
Hence, from the given diagram we get that,
Sinθ = h/AD.
So, we will have that v^2=u^2-2gS. Where S is ADSinθ
Hence, VB^2 = vA^2 - 2gSinθ AD.
On solving we will get that VB^2=(√136)^2 - 2*10Sin 37*(3/Sin37) which on solving we will get vB=√76 m/s.
Hence, finally again using Newton's law of motion we will get the velocity of the point B as Vb^2 = vB^2 -2aX where X is the length of the section which is 3m. So, Vb²= (√76)² -2*10*3 which on solving we will get Vb=4 m/s.
Answer:
From the given question we get that the length is L = 3m and the height given is h=3m and the speed of the block when it passes the point A is vA = √136m/s with the angle of inclination of θ = 37°.
From the given question we get that the length is L = 3m and the height given is h=3m and the speed of the block when it passes the point A is vA = √136m/s with the angle of inclination of θ = 37°.Again for the block we will have that mgsinθ + μ*mgcosθ = ma(where μ is the coefficient of friction)
From the given question we get that the length is L = 3m and the height given is h=3m and the speed of the block when it passes the point A is vA = √136m/s with the angle of inclination of θ = 37°.Again for the block we will have that mgsinθ + μ*mgcosθ = ma(where μ is the coefficient of friction)Thus, the acceleration a = g(sin37 + cos 37)
From the given question we get that the length is L = 3m and the height given is h=3m and the speed of the block when it passes the point A is vA = √136m/s with the angle of inclination of θ = 37°.Again for the block we will have that mgsinθ + μ*mgcosθ = ma(where μ is the coefficient of friction)Thus, the acceleration a = g(sin37 + cos 37)Which on solving we will get a = 10 m/s^2
From the given question we get that the length is L = 3m and the height given is h=3m and the speed of the block when it passes the point A is vA = √136m/s with the angle of inclination of θ = 37°.Again for the block we will have that mgsinθ + μ*mgcosθ = ma(where μ is the coefficient of friction)Thus, the acceleration a = g(sin37 + cos 37)Which on solving we will get a = 10 m/s^2Hence, from the given diagram we get that,
From the given question we get that the length is L = 3m and the height given is h=3m and the speed of the block when it passes the point A is vA = √136m/s with the angle of inclination of θ = 37°.Again for the block we will have that mgsinθ + μ*mgcosθ = ma(where μ is the coefficient of friction)Thus, the acceleration a = g(sin37 + cos 37)Which on solving we will get a = 10 m/s^2Hence, from the given diagram we get that,Sinθ = h/AD.
From the given question we get that the length is L = 3m and the height given is h=3m and the speed of the block when it passes the point A is vA = √136m/s with the angle of inclination of θ = 37°.Again for the block we will have that mgsinθ + μ*mgcosθ = ma(where μ is the coefficient of friction)Thus, the acceleration a = g(sin37 + cos 37)Which on solving we will get a = 10 m/s^2Hence, from the given diagram we get that,Sinθ = h/AD.So, we will have that v^2=u^2-2gS. Where S is ADSinθ
From the given question we get that the length is L = 3m and the height given is h=3m and the speed of the block when it passes the point A is vA = √136m/s with the angle of inclination of θ = 37°.Again for the block we will have that mgsinθ + μ*mgcosθ = ma(where μ is the coefficient of friction)Thus, the acceleration a = g(sin37 + cos 37)Which on solving we will get a = 10 m/s^2Hence, from the given diagram we get that,Sinθ = h/AD.So, we will have that v^2=u^2-2gS. Where S is ADSinθHence, VB^2 = vA^2 - 2gSinθ AD.
From the given question we get that the length is L = 3m and the height given is h=3m and the speed of the block when it passes the point A is vA = √136m/s with the angle of inclination of θ = 37°.Again for the block we will have that mgsinθ + μ*mgcosθ = ma(where μ is the coefficient of friction)Thus, the acceleration a = g(sin37 + cos 37)Which on solving we will get a = 10 m/s^2Hence, from the given diagram we get that,Sinθ = h/AD.So, we will have that v^2=u^2-2gS. Where S is ADSinθHence, VB^2 = vA^2 - 2gSinθ AD.On solving we will get that VB^2=(√136)^2 - 2*10Sin 37*(3/Sin37) which on solving we will get vB=√76 m/s.