Respuesta :
Answer:
a) Wf = -21.936 J
b) Wg = 16.18 J
c) Wn = 0 J
d) W = -5.756 J
e) v2 = 1.65 m/s
Explanation:
Given:
- The mass of the package m = 5.20 kg
- The inclination of ramp θ = 13°
- The coefficient of kinetic friction mk = 0.313
- The package slides down a distance of s = 1.41 m
- The initial velocity of the package v1 = 2.22 m/s
Find:
(a) the work done on the package by friction;
(b) the work done on the package by gravity;
(c) the work done on the package by the normal force;
(d) the total work done on the package.
(e) If the package has a speed of 2.22 m>s at the top of the ramp, what is its speed after it has slid 1.41 m down the ramp?
Solution:
- Apply Equilibrium conditions perpendicular to the surface:
N - m*g*cos(θ) = 0
N =m*g*cos(θ)
Where, N is the normal contact force exerted by the surface on the package.
- The frictional force Ff is given by:
Ff = mk*N
Ff = -mk*m*g*cos(θ)
- The work done Wf by frictional force is given by:
Wf = Ff*s
Wf = -s*mk*m*g*cos(θ)
Plug in values:
Wf = -1.41*0.313*5.2*9.81*cos(13)
Wf = -21.936 J
- The work done by the contact force Wn is given by:
Wn = s_y*N
Since, the equilibrium conditions are maintained perpendicular to surface and no distance is traveled; hence, s_y = 0
Wn = 0 J
- The total work done by gravity Wg is given by moving the vertical distance h from start to finish:
Wg = m*g*h
Wg = m*g*s*sin(θ)
Wg = 5.2*9.81*1.41*sin(13)
Wg = 16.18 J
- So, the total work done W is given by the sum:
W = Wf + Wn + Wg
W = -21.936 + 0 + 16.18
W = -5.756 J
- Construct an energy balance for the two states at top and bottom of the ramp:
0.5*m*v1^2 + W = 0.5*m*v2^2
2*(v1^2 + W/m) = v2^2
v2 = sqrt ( (v1^2 + 2*W/m) )
Plug in values:
v2 = sqrt ( (2.22^2 -5.756*2/5.2) )
v2 = sqrt ( 2.71455 )
v2 = 1.65 m/s
