A force F is applied in horizontal to a 10 kg block. The block moves at a constant speed of 2 m/s across a horizontal surface. The coefficient of kinetic frison between the block and the surface is 0.5. The work done by
the force F in 1.5 minutes is:

Part f - example: finding two forces (part i) two dimensional dynamics often involves solving for two unknown quantities in two separate equations describing the total force. the block in (figure 1) has a mass m=10kg and is being pulled by a force f on a table with coefficient of static friction îľs=0.3. four forces act on it: the applied force f (directed î¸=30â above the horizontal). the force of gravity fg=mg (directly down, where g=9.8m/s2). the normal force n (directly up). the force of static friction fs (directly left, opposing any potential motion). if we want to find the size of the force necessary to just barely overcome static friction (in which case fs=îľsn), we use the condition that the sum of the forces in both directions must be 0. using some basic trigonometry, we can write this condition out for the forces in both the horizontal and vertical directions, respectively, as: fcosî¸â’îľsn=0 fsinî¸+nâ’mg=0 in order to find the magnitude of force f, we have to solve a system of two equations with both f and the normal force n unknown. use the methods we have learned to find an expression for f in terms of m, g, î¸, and îľs (no n).

The boy of mass 30 kg is at r1= 2 meters from the fulcrum. of the girl is 45 kg, at what r2 must she sit so that they are balanced ?

If the wheels lock when braking suddenly the vehicle: a: loses traction b: lose steering wheel ability c: gain speed slightly d: gain steering ability