A 5.2 kg block is pushed 4.8 m up a rough 39° inclined plane by a horizontal force of 80 N. If the initial spe?
A 5.2 kg block is pushed 4.8 m up a rough 39° inclined plane by a horizontal force of 80 N. If the initial speed of the block is 2.4 m/s up the plane and a constant kinetic friction force of 21 N opposes the motion, calculate the following.
a) the work done by the horizontal force? (answer in J)
b) the final kinetic energy of the block? (answeer in J)
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a) The work done by the horizontal force = F*d*cos(theta) Here: F = 80 N, d = 4.8 m (displacement), theta = 39 degree (angle between the force and the direction of motion). Therefore
W = (80)(4.8)(cos39) = 298 J
b) To find the final kinetic energy we will find the: (1) net force, (2) acceleration, (3) final speed, (4) final kinetic energy
(1) The net force = parallel component of pushing force(up the incline) - force of friction("-" b/c the force points down the incline) - parallel component of the force of gravity("-" b/c the force points down the incline) = (80N)(cos39) - (21N) - (5.2kg)(9.8m/s^2)(sin39) = 62N - 21N - 32N = 9 N
NOTE: The perpendicular components of the above forces do not change the kinetic energy of the block
(2) The net force accelerates the block at a = (9N)/(5.2kg) = 1.73 m/s^2
(3) Now we can find the final speed of the block: (Vf)^2 = (Vo)^2 + 2*a*d.
Therefore V(final) = 4.73 m/s.
(4) The final kinetic energy = 1/2mv^2 = 58 J
beginning up, once you observed "horizontal rigidity" i assume you mean, "horizontal to the plane." kinetic potential is a million/2mv^2 = 12.a million joules paintings is F * d, so paintings carried out on the block via the horizontal rigidity is 525 J paintings carried out via the frictional rigidity (additionally horizontal to the plane) is then 21N * 7m = 147 J to locate paintings carried out via gravity, could get the ingredient of the load of the field that acts down the plane, (sin 37)(5kg)(9.8m/s^2) = (sin 37)(49N) = 29.5 N the plane is, returned, 7m, so paintings carried out is 206.4 J the paintings carried out via the traditional rigidity (cos 37)(coefficient of friction)(49N), is a similar by using fact the frictional rigidity, or 147 J. additionally, for the latter area of this question, it rather is definitely the "comparable," no longer merely a similar huge style, so which you do no longer count huge style it two times. by using fact the forces parallel to the plane do no longer stability, there's a internet rigidity interior the upward course alongside the plane. rigidity on field alongside the plane (up) - frictional rigidity - gravity appearing down the plane = internet rigidity on field in keeping with the plane. yet, given which you have already calculated the potential, you could merely upload and subtract potential somewhat of going returned to forces (paintings = potential). So, 525J - 147J - 206.4J = 171.6J, that's the internet potential extra to the preliminary KE. on condition that preliminary became 12.1J, including that supplies you a entire very final KE of 183.7J a greater suitable (no longer asked) question is, how briskly is the field now shifting? that's 8.fifty seven m/s