Problem 2: Newton's Laws M A 1.27-kg mass M slides down a plane inclined at an angle e = 55° to the horizontal. There is a 2.66•N frictional force on the mass. Take the x•axis along the plane in the generally downward direction and take the y•axis normal to the plane in the generally upward direction, as shown in the figure. (a) Draw a free body diagram of the mass. Be sure to label all forces clearly and to place the angle 0 in your diagram. mgy =mgsin 55. mgy =mgcos550 0 = FN-mgy O = FM-mgcos5s FN =mg cos55 m = \.27kg O = 55° (b) Find the acceleration of the mass M down the plane in the x•direction. mgx: mgsins5= ma gsin55 = a a-8.04 misz- FF =ma 5.38m/s² ax= a= 5.38M/s2 (C) What is the acceleration of the mass M in the y•direction? Explain your answer. a =5,63 m/s 2 (d) Find the magnitude of the normal force acting on the mass M . 9, COS55 = a ay = 2 5,63 m/s FN=mgcos55 FN = (1.27kg)(9.81mis) (cos(55)) FM=7,15N 7.15N n = (e) Find the coefficient of kinetic friction between the mass and the plane. yhgsine-Myg coso = vha I (sine -MOSB)=ax M ="a+gsin M-4.6? Cose Hk=
Problem 2: Newton's Laws M A 1.27-kg mass M slides down a plane inclined at an angle e = 55° to the horizontal. There is a 2.66•N frictional force on the mass. Take the x•axis along the plane in the generally downward direction and take the y•axis normal to the plane in the generally upward direction, as shown in the figure. (a) Draw a free body diagram of the mass. Be sure to label all forces clearly and to place the angle 0 in your diagram. mgy =mgsin 55. mgy =mgcos550 0 = FN-mgy O = FM-mgcos5s FN =mg cos55 m = \.27kg O = 55° (b) Find the acceleration of the mass M down the plane in the x•direction. mgx: mgsins5= ma gsin55 = a a-8.04 misz- FF =ma 5.38m/s² ax= a= 5.38M/s2 (C) What is the acceleration of the mass M in the y•direction? Explain your answer. a =5,63 m/s 2 (d) Find the magnitude of the normal force acting on the mass M . 9, COS55 = a ay = 2 5,63 m/s FN=mgcos55 FN = (1.27kg)(9.81mis) (cos(55)) FM=7,15N 7.15N n = (e) Find the coefficient of kinetic friction between the mass and the plane. yhgsine-Myg coso = vha I (sine -MOSB)=ax M ="a+gsin M-4.6? Cose Hk=
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Transcribed Image Text:Section II: Problems
Problem 2: Newton's Laws
Page 4
A 1.27•kg mass M slides down a plane inclined at an angle e = 55° to the horizontal.
There is a 2.66•N frictional force on the mass. Take the x•axis along the plane in the
generally downward direction and take the y•axis normal to the plane in the generally
upward direction, as shown in the figure.
%3D
(2) Draw a free body diagram of the mass. Be sure to label all forces clearly and to
place the angle e in your diagram.
mgy =mgsin 55.
moy =mgcos5s.
0 = FN-mgy
O = FM-mgcos55
FN=mg cus55
m= 1.27kg
O = 55°
(b) Find the acceleration of the mass M down the plane in the x•direction.
mgx=ma
igsins5 = ma
gsin55 = a
a-8.04 misz- Ff
%3D
%3D
5.38m/5²
ax=
a= 5.38MIs2
(c) What is the acceleration of the mass M in the y•direction? Explain your answer.
%3D
9u COS55 = a
ay=_ 5,63m/s 2
a =5,63 m/s
O Find the magnitude of the normal force acting on the mass M
ay =
FN=mgcos55
FN = (1.27kg)(9.81mis)(cos(55))
FM=7.15M
7.15 N
n =
(e) Find the coefficient of kinetic friction between the mass and the plane.
yigsine-Mygcose = rha
I (sine -MCOSB)=x
M ="a+9sinA M=4.6?
Cose
Hk =
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