d = Vo*t + 0.5at^2 = 11.1m,
0 + 0.5a*(2.4)^2 = 11.1,
2.88a = 11.1,
a = 3.85m/s^2.
Fn = 72 - fcos21.4 = ma,
-72 - Fcos21.4 = 8*3.85 = 30.6,
-Fcos21.4 = 30.6 + 72 = 102.6,
-F =110.2N
F = -110.2N. t0 the left.
NOTE: The sine of 21.4 was not used,
because it does not contribute to hor.
movement.
A mass of 8 kg lies on a horizontal, frictionless floor. A force of 72 Newtons pushes to the left (negative x direction) with a force of 72 Newtons. Another force of unknown magnitude pushes the mass in a direction of 21.4 degrees above the positive x axis. The mass is originally at rest before these forces are applied and 2.4 seconds after the forces have been applied, the mass has moved to the left a distance of 11.1 meters. What is the magnitude of the unknown force in Newtons?
1 answer