Question

Determine the distance s when the block stops Note that the origin of s is defined as the beginning of the spring. Express your answer to three significant figures and include the appropriate units. When s 0.4 m, the spring is unstretched and a block of mass 10kg has a speed of v- 5.5 m/s moving down a plane that is inclined at an angle of 8-35 degrees as shown. The plane and the block has a kinetic friction coefficient of 0.25, while the spring constant is 225 N/m. A force F, of constant magnitude 80 N acts in a direction at an angle of B- 20 degrees with respect to the plane (Figure 1) Value Units Submit Part B igure 1 of 1 What is the speed of the block when the distance s1.299 m? alue nits Submit est Ans <Return to Assignment Provide Feedback

Answer 1

We can solve this problem using the Energy Balance method

Assume Origin as Shown in fig:

Given Data: Mass of Block (m) = 10kg Unstretched Length of spring (s) = 0.4m Spring Constant (k) = 225 N/m The coefficient of Kinetic friction�(mu)�= 0.25 The speed of block at the origin (v) = 5.5 m/s The angle of inclination of the plane��theta = 35 degree External Force acting (F) = 80 N The angle of external force with respect to the plane�beta = 20 degree PART A The weight of Spring W = 10 * 9.81 = 98.1 N The angle between the normal reaction (N) and weight (W) =�theta = 35 degree �(N is perpendicular to the base of� the triangle and W is perpendicular to the hypotenuse of the triangle. So the angle between W and N will�be the same as the angle made by the base and hypotenuse of the triangle) Normal reaction N = W cos 35o�= 98.1 *cos 35o�= 80.3588 N \r\n

Given Data :

Mass of Block (m) = 10kg

Unstretched Length of spring (s)= 0.4m

Spring Constant (k) = 225 N/m

The coefficient of Kinetic friction $(\mu)$ = 0.25

The speed of block at the origin (v)= 5.5 m/s

The angle of inclination of the plane   $\theta = 35^{o}$

External Force acting (F) = 80 N

The angle of external force with respect to the plane $\beta = 20^{o}$

PART A

The weight of Spring W =10 * 9.81 = 98.1 N

The angle between the normal reaction (N) and weight (W) = $\theta = 35^{o}$ (N is perpendicular to the base of the triangle and W is perpendicular to the hypotenuse of the triangle. So the angle between W and N will be the same as the angle made by the base and hypotenuse of the triangle )

Normal reaction N = W cos 35^o = 98.1 *cos 35^o = 80.3588 N

Component of weight acting along x-axis = W sin 35^o = 98.1 *sin 35^o = 56.2678 N

At origin:

• Kinematic friction (f) = $\mu$ x N = 0.25 x 80.3588 = - 20.0897 N ( -ve implies force acts in -ve x direction )
• Kinetic Energy of Block (K.E) = 1/2 mv² = 1/2 x 10 x 5.5² =151.25 Joules
• Work done by external forces (F,f) = 0 (No displacement has occurred in the direction of forces as the body is at the origin)
• The potential energy of spring =0 (the spring is not stretched)

  

Therefore total Energy of the system when the block is at origin = 151.25 Joules ----(1)

Let x be the distance from origin where the body comes to stop.

At that point, the total energy balance equation is as follows

Initial KE (1) = Potential Energy of Spring ( due to extension 'x') + Work done by external forces (F and f, w sin $\theta$ ) +Final KE of the block (=0 as the block comes to stop)

Potential Energy stored in the Spring due to extension = 1/2 kx² = 1/2 * 225 *x ² = - 112.5x² J (Negative is given as the force and displacement is in opposite direction)---(2)

Work done by friction = f * x = - (20.0897 * x) J= -20.0897x J (Negative is given as the force and displacement is in opposite direction)----(3)

Component of Force F in the direction of displacement of the block = F cos $\beta$ = F cos 20^o = 80 * cos 20^o = 75.1754 N

Work done by the component of Force F in the direction of displacement of the block = 75.1754 * x J (4)

Work done by the component of Weight in the direction of displacement of the block = 56.2678 * x J ---- (5)

Work done in support of motion of Block are (1),(4),(5)

Work done againest motion of Block are (2),(3)

Equating the the energy

112.5x² + 20.0897x = 151.25 + 75.1754x + 56.2678x

112.5x² -111.3535 x - 151.25 =0 -----(6)

Solving equation (6) , we get x =1.7556 or x = -0.765 . As the negative values doesnt make sense, we ignore it and the value of x = 1.7556m

Therefore the distance block stops, meansured from the origin of spring = s+x = 0.4 +1.7556 = 2.16 m.

PART B

at s= 1.299

x = 0.899

At that point, the total energy balance equation is as follows

Potential Energy of Spring ( due to extension 'x') + Work done by external forces (F and f, w sin $\theta$ ) + Initial KE (1) - KE of the block at that point =0

Potential Energy stored in the Spring due to extension = 1/2 kx² = 1/2 * 225 *0.899 ² = - 90.9226J (Negative is given as the force and displacement is in opposite direction)---(7)

Work done by friction = f * x = - (20.0897 * 0.899) J= -18.0787 J (Negative is given as the force and displacement is in opposite direction)----(8)

Component of Force F in the direction of displacement of the block = F cos $\beta$ = F cos 20^o = 80 * cos 20^o = 75.1754 N

Work done by the component of Force F in the direction of displacement of the block = 75.1754 * 0.899= 67.58268 J (9)

Work done by the component of Weight in the direction of displacement of the block = 56.2678 * 0.899 = 50.58475 J ---- (10)

Kinetic energy of block at point (x =0.899) = 1/2 mv² = 1/2 * 10* v² = 5v² J -----(11)

Work done in support of motion of Block are (1),(9),(10),(11)

Work done againest motion of Block are (7),(8)

Equating the the energy

90.9226 + 18.0787 +5v² = 151.25 + 67.58268 + 50.58475 ----(12)

Solving equation (12)

We get v = 5.664 m/s