MasteringPhysics: Work and Kinetic Energy- Google Chrome ecure https://session.m
ID: 1780524 • Letter: M
Question
MasteringPhysics: Work and Kinetic Energy- Google Chrome ecure https://session.masteringphysics.com/myct/itemView?assignmentProblemID 87499287koffset-prev ork and Kinetic Energy Problem 9.45 e previous| 2 roblem 9.45 Part A Susan's 10.0 kg baby brother Paul sits on a mat Susan pulls the mat across the floor using a rope that is angled 30 above the floor The tension is a constant 28 0 N and the coefficient of friction is 0.170 Use work and energy to find Paurs speed after being pulled 3 20 Express your answer with the appropriate units. HA Value Units Submit My Answers Give Up
Explanation / Answer
If mass m is pulled horizontally by a rope with tension T at angle to the horizontal, the vertical component of T reduces the effect of the weight, so that normal force N is given by
N = mg - Tsin ...........................................
Horizontal friction force Fr opposing motion is given by
Fr = N = (mg - Tsin ) ...........................................
In this example, the horizontal component of T exceeds the opposing horizontal friction force
Excess force available for acceleration is F = Tcos - (mg - Tsin ) ...(3)
[Note: We could use acceleration a = excess force/mass = [Tcos - (mg - Tsin )]/m
and the equation of motion v2 = 2aS at this stage, but we are directed to use work and energy to find final speed v. This is equivalent.]
The kinetic energy of the mass with final speed v at distance S from the start is provided by the work done by excess force F moving that distance S. In other words,
(1/2)mv2 = F*S or
v2 = 2(F/m)S
v2 = 2[Tcos - (mg - Tsin )]/m *S
Tension T = 28 N, cos 30 = (3)/2, sin 30 = 1/2, = 0.17, m = 10 kg, g = 9.81 m/s^2, sin 30 = 1/2, S = 3.2 m
v2 = 2[14(3)-0.17(98.1-14)]/(10) * 3.2
v = 2.52 m/s