Assume first that the 50kg. moves up the plane.

The forces on the 50kg. are:

(a) its weight vertically downward;

(b) friction down and parallel to the plane (opposing motion);

(c) the reaction (call this R) of the plane perpendicular to the plane;

(d) the tension (call this T) in the string.

The forces on the 10kg. are:

(a) the tension T upward;

(b) its weight downward.

Let:

the downward acceleration of the 10kg. be a m/sec^2;

the required distance be s metre;

the acceleration due to gravity be g m/sec^2.

For the 10kg:

(10 – T)g = 10a …(1)

For the 50kg, bearing in mind that there is no motion perpendicular to the plane:

50cos(53) = R …(2)

(T – 50sin(53) – 0.25R)g = 50a …(3)

Using the standard uniform acceleration equation

s = ut + (1/2)at^2

for the 10kg.:

s = 2 + a / 2 …(4)

From (1):

T = 10(g – a) / g

Substitute for T and for R from (2) in (3):

10(g – a) – (50sin(53) + 0.25*50cos(53))g = 50a …(5)

Adding 10a:

10g – (50sin(53) + 0.25*50cos(53))g = 60a

a = (10 – 50sin(53) – 0.25*50cos(53))g/60 …(6)

a = -6.12m/sec.

The assumption that the 50kg. moves up the plane is therefore incorrect.

The friction and the acceleration therefore need to be reversed.

The revised equations are:

(T – 10)g = 10a …(1)

50cos(53) = R …(2)

(50sin(53) – 0.25R – T)g = 50a …(3)

s = 2 + a / 2 …(4)

yielding:

a = (50sin(53) – 0.25*50cos(53) – 10)g / 60

= 3.66m/sec.

Hence from (4):

s = 2 + 1.83

= 3.83sec.

a) The wagon strikes at consistent velocity. consequently, the strain of the pulling is comparable with the frictional tension. F = ma consistent velocity, F cos 30 = 30 F = 30 / (cos 30) F = 34.641 N b) Now the wagon hastens at 0.4 m/s^2, there is an unbalanced tension. it is because of the fact the strain of pulling is extra advantageous than frictional tension. consequently, tension of pulling – Frictional tension = Unbalanced tension. tension of pulling = Unbalanced tension + Frictional tension tension of pulling = 20 kg (0.4 m/s^2) + 30 N tension of pulling = 38 N tension of pulling = F cos 30 = 38 F = 38 / (cos 30) F = 40 3.87 N