Fg = M*g = 4 * 9.8 = 39.2 N. = Force of gravity.
Fp = 39.2*sin30 = 19.6 N. = Force parallel to the incline.
Fn = 39.2*Cos30 = 33.9 N. = Normal force
a. W = Fg*d = 39.2 * 10 = 392 J.
b. W = Fn*d = 33.9 * 10 = 339 J.
c.
d. h = d*sin30 = 10*sin30 = 5 m.
KE = PE = Mg * h = 39.2 * 5 = 196 J.
Starting from rest, a 4-kg block slides 10 m down a frictionless 30 degree incline. Determine the work done on the block by
a) The force of gravity
b) The normal force
c) All of the forces (the net force) on the block
d) Find the Kinetic energy of the block at the end of the 10 m slide.
2 answers
A)w=mgcos30
W=4x9.8xcos30x10
=196
B) w=mgsin30
W=4x9.8xsin30x10
W=33.95j
C)net force
196+33.95
Net woek=229.95
D)ke=pe
Ke=1/2mv²
V is unknown
V²=u²+2ad
V²=0²+2x9.8×10
V=14m/s
Ke=1/2×4×14
Ke=28joules
W=4x9.8xcos30x10
=196
B) w=mgsin30
W=4x9.8xsin30x10
W=33.95j
C)net force
196+33.95
Net woek=229.95
D)ke=pe
Ke=1/2mv²
V is unknown
V²=u²+2ad
V²=0²+2x9.8×10
V=14m/s
Ke=1/2×4×14
Ke=28joules