Physics, asked by Anonymous, 10 months ago

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.

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Answers

Answered by AneesKakar
1

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.

Answered by mustafa3952
0

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.

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.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.

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